Chuyên đề: Các bài tập nâng cao trong rút gọn đa thức

A. Lý thuyết

  • Nắm vững các dạng rút gọn, các hằng đẳng thức, quy đồng đặt nhân tử chung,...

B. Bài tập

I. Bài tập minh họa

Câu 1.  Cho biểu thức P=(ab+ba+1)(1a−1b)2a2b2+b2a2−(ab+ba)P=\frac{(\frac{a}{b}+\frac{b}{a}+1){{(\frac{1}{a}-\frac{1}{b})}^{2}}}{\frac{{{a}^{2}}}{{{b}^{2}}}+\frac{{{b}^{2}}}{{{a}^{2}}}-(\frac{a}{b}+\frac{b}{a})} với a > 0, b > 0, a ≠ b.

  1. Chứng minh P=1abP=\frac{1}{ab}
  2. Giả sử a, b thay đổi sao cho 4a+b+ab=14a+b+\sqrt{ab}=1. Tìm giá trị nhỏ nhất của P.

Giải

Ta có:

P=(ab+ba+1)(1a−1b)2a2b2+b2a2−(ab+ba)=(a2ab+b2ab+abab)(a−bab)2a4a2b2+b4a2b2−(a3ba2b2+ab3a2b2)=(a3−b3)(a−b)a3b3(a3−b3)(a−b)a2b2=1abP=\frac{(\frac{a}{b}+\frac{b}{a}+1){{(\frac{1}{a}-\frac{1}{b})}^{2}}}{\frac{{{a}^{2}}}{{{b}^{2}}}+\frac{{{b}^{2}}}{{{a}^{2}}}-(\frac{a}{b}+\frac{b}{a})}=\frac{\left( \frac{{{a}^{2}}}{ab}+\frac{{{b}^{2}}}{ab}+\frac{ab}{ab} \right){{\left( \frac{a-b}{ab} \right)}^{2}}}{\frac{{{a}^{4}}}{{{a}^{2}}{{b}^{2}}}+\frac{{{b}^{4}}}{{{a}^{2}}{{b}^{2}}}-\left( \frac{{{a}^{3}}b}{{{a}^{2}}{{b}^{2}}}+\frac{a{{b}^{3}}}{{{a}^{2}}{{b}^{2}}} \right)}=\frac{\frac{\left( {{a}^{3}}-{{b}^{3}} \right)\left( a-b \right)}{{{a}^{3}}{{b}^{3}}}}{\frac{\left( {{a}^{3}}-{{b}^{3}} \right)\left( a-b \right)}{{{a}^{2}}{{b}^{2}}}}=\frac{1}{ab}

Câu 2.

Cho hai số thực a , b thỏa điều kiện ab = 1, a +b ¹ 0 . Tính giá trị của biểu thức:

      P=1(a+b)3(1a3+1b3)+3(a+b)4(1a2+1b2)+6(a+b)5(1a+1b)P=\frac{1}{{{(a+b)}^{3}}}(\frac{1}{{{a}^{3}}}+\frac{1}{{{b}^{3}}})+\frac{3}{{{(a+b)}^{4}}}(\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}})+\frac{6}{{{(a+b)}^{5}}}(\frac{1}{a}+\frac{1}{b})

Giải

Với ab = 1 , a + b ¹ 0, ta có:

P=1(a+b)3(1a3+1b3)+3(a+b)4(1a2+1b2)+6(a+b)5(1a+1b)P=\frac{1}{{{(a+b)}^{3}}}(\frac{1}{{{a}^{3}}}+\frac{1}{{{b}^{3}}})+\frac{3}{{{(a+b)}^{4}}}(\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}})+\frac{6}{{{(a+b)}^{5}}}(\frac{1}{a}+\frac{1}{b})=a3+b3(a+b)3+3(a2+b2)(a+b)4+6(a+b)(a+b)5=a2+b2−1(a+b)2+3(a2+b2)(a+b)4+6(a+b)(a+b)5=\frac{{{a}^{3}}+{{b}^{3}}}{{{\left( a+b \right)}^{3}}}+\frac{3\left( {{a}^{2}}+{{b}^{2}} \right)}{{{\left( a+b \right)}^{4}}}+\frac{6\left( a+b \right)}{{{\left( a+b \right)}^{5}}}=\frac{{{a}^{2}}+{{b}^{2}}-1}{{{\left( a+b \right)}^{2}}}+\frac{3\left( {{a}^{2}}+{{b}^{2}} \right)}{{{\left( a+b \right)}^{4}}}+\frac{6\left( a+b \right)}{{{\left( a+b \right)}^{5}}}=(a2+b2−1)(a+b)2+3(a2+b2)+6(a+b)4=(a2+b2−1)(a2+b2+2)+3(a2+b2)+6(a+b)4=\frac{\left( {{a}^{2}}+{{b}^{2}}-1 \right){{\left( a+b \right)}^{2}}+3\left( {{a}^{2}}+{{b}^{2}} \right)+6}{{{\left( a+b \right)}^{4}}}=\frac{\left( {{a}^{2}}+{{b}^{2}}-1 \right)\left( {{a}^{2}}+{{b}^{2}}+2 \right)+3\left( {{a}^{2}}+{{b}^{2}} \right)+6}{{{\left( a+b \right)}^{4}}}=(a2+b2+2)2(a+b)4=(a2+b2+2ab)2(a+b)4=1=\frac{{{\left( {{a}^{2}}+{{b}^{2}}+2 \right)}^{2}}}{{{\left( a+b \right)}^{4}}}=\frac{{{\left( {{a}^{2}}+{{b}^{2}}+2ab \right)}^{2}}}{{{\left( a+b \right)}^{4}}}=1

Vậy P = 1, với ab = 1 , a+b ¹ 0.

Câu 3.

Cho các số thực dương a, b.Chứng minh rằng

                        (a−b)3(a−b)3−bb+2aaaa−bb+3a+3abb−a=0\frac{\frac{{{(a-b)}^{3}}}{{{(\sqrt{a}-\sqrt{b})}^{3}}}-b\sqrt{b}+2a\sqrt{a}}{a\sqrt{a}-b\sqrt{b}}+\frac{3a+3\sqrt{ab}}{b-a}=0

Giải

Q=(a−b)3(a−b)3−bb+2aaaa−bb+3a+3abb−aQ=\frac{\frac{{{(a-b)}^{3}}}{{{(\sqrt{a}-\sqrt{b})}^{3}}}-b\sqrt{b}+2a\sqrt{a}}{a\sqrt{a}-b\sqrt{b}}+\frac{3a+3\sqrt{ab}}{b-a}=(a−b)3.(a+b)3(a−b)3−bb+2aa(a−b)(a+ab+b)−3a(a+b)(a−b)(a+b)=\frac{\frac{{{(\sqrt{a}-\sqrt{b})}^{3}}.{{(\sqrt{a}+\sqrt{b})}^{3}}}{{{(\sqrt{a}-\sqrt{b})}^{3}}}-b\sqrt{b}+2a\sqrt{a}}{(\sqrt{a}-\sqrt{b})(a+\sqrt{ab}+b)}-\frac{3\sqrt{a}(\sqrt{a}+\sqrt{b})}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}=aa+3ab+3ba+bb+2aa(a−b)(a+ab+b)−3aa−b=\frac{a\sqrt{a}+3a\sqrt{b}+3b\sqrt{a}+b\sqrt{b}+2a\sqrt{a}}{(\sqrt{a}-\sqrt{b})(a+\sqrt{ab}+b)}-\frac{3\sqrt{a}}{\sqrt{a}-\sqrt{b}}=3aa+3ab+3ba−3aa−3ab−3ba(a−b)(a+ab+b)=\frac{3a\sqrt{a}+3a\sqrt{b}+3b\sqrt{a}-3a\sqrt{a}-3a\sqrt{b}-3b\sqrt{a}}{(\sqrt{a}-\sqrt{b})(a+\sqrt{ab}+b)}=0(DPCM)=0(DPCM)

Câu 4.Cho các số dương x, y, z thỏa mãn các điều kiện x + y + z = 2 và x2 + y2 + z2 = 2.

Chứng minh rằng biểu thức sau không phụ thuộc vào x, y, z:

                        P=x(1+y2)(1+z2)1+x2+y(1+x2)(1+z2)1+y2+z(1+y2)(1+x2)1+z2P=x\sqrt{\frac{\left( 1+{{y}^{2}} \right)\left( 1+{{z}^{2}} \right)}{1+{{x}^{2}}}}+y\sqrt{\frac{\left( 1+{{x}^{2}} \right)\left( 1+{{z}^{2}} \right)}{1+{{y}^{2}}}}+z\sqrt{\frac{\left( 1+{{y}^{2}} \right)\left( 1+{{x}^{2}} \right)}{1+{{z}^{2}}}}

Giải

Xét (x+y+z)2=x2+y2+z2+2(xy+xz+yz)⇒xy+yz+xz=(x+y+z)2−(x2+y2+z2)2{{\left( x+y+z \right)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2\left( xy+x\text{z}+yz \right)\Rightarrow xy+yz+x\text{z}=\frac{{{\left( x+y+z \right)}^{2}}-\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)}{2}

Thay x + y + z = 2 và x2 + y2 + z2 = 2 ta có xy + yz + zx = 1.

Thay 1 = xy + yz + zx ta có:

x(1+y2)(1+z2)1+x2=x(xy+yz+zx+y2)(xy+yz+zx+z2)xy+yz+zx+x2=x(y+z)(y+x)(z+y)(z+x)(x+y)(x+z)=x(y+z)x\sqrt{\frac{\left( 1+{{y}^{2}} \right)\left( 1+{{z}^{2}} \right)}{1+{{x}^{2}}}}=x\sqrt{\frac{\left( xy+yz+z\text{x}+{{y}^{2}} \right)\left( xy+yz+z\text{x}+{{z}^{2}} \right)}{xy+yz+z\text{x}+{{x}^{2}}}}=x\sqrt{\frac{\left( y+z \right)\left( y+x \right)\left( z+y \right)\left( z+x \right)}{\left( x+y \right)\left( x+z \right)}}=x\left( y+z \right)

Tương tự ta có:

y(1+z2)(1+x2)1+y2=y(z+x)y\sqrt{\frac{\left( 1+{{z}^{2}} \right)\left( 1+{{x}^{2}} \right)}{1+{{y}^{2}}}}=y\left( z+x \right)

z(1+y2)(1+x2)1+z2=z(y+x)z\sqrt{\frac{\left( 1+{{y}^{2}} \right)\left( 1+{{x}^{2}} \right)}{1+{{z}^{2}}}}=z\left( y+x \right)

Cộng từng vế của ba đẳng thức trên ta có

P=xy+xz+yz+xy+zx+zy=2(xy+yz+zx)=2P=xy+x\text{z}+yz+xy+z\text{x}+zy=2\left( xy+yz+z\text{x} \right)=2

Vậy biểu thức P không phụ thuộc vào x, y, z.

Câu 5.

Cho 3 số thực x, y, z thỏa mãn điều kiện: x + y + z = 0 và xyz ≠ 0.

Tính giá trị biểu thức P=x2y2+z2−x2+y2z2+x2−y2+z2x2+y2−z2P=\frac{{{x}^{2}}}{{{y}^{2}}+{{z}^{2}}-{{x}^{2}}}+\frac{{{y}^{2}}}{{{z}^{2}}+{{x}^{2}}-{{y}^{2}}}+\frac{{{z}^{2}}}{{{x}^{2}}+{{y}^{2}}-{{z}^{2}}}

Giải

Ta có

x+y+z=0⇔(y+z)2=(−x)2⇔y2+z2−x2=−2yzx+y+z=0\Leftrightarrow {{\left( y+z \right)}^{2}}={{\left( -x \right)}^{2}}\Leftrightarrow {{y}^{2}}+{{z}^{2}}-{{x}^{2}}=-2yz

Tương tự:

z2+x2−y2=−2zx{{z}^{2}}+{{x}^{2}}-{{y}^{2}}=-2\text{zx}

x2+y2−z2=−2yz{{x}^{2}}+{{y}^{2}}-{{z}^{2}}=-2yz

P=x2−2yz+y2−2zx+z2−2yz=x3+y3+z3−2xyzP=\frac{{{x}^{2}}}{-2yz}+\frac{{{y}^{2}}}{-2\text{zx}}+\frac{{{z}^{2}}}{-2yz}=\frac{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}}{-2\text{x}yz}

Mà

x3+y3+z3=(x+y)3−3x2y−3xy2+z3=(−z)3−3xy(x+y)+z3=3xy⇒P=3xyz−2xyz=−32{{x}^{3}}+{{y}^{3}}+{{z}^{3}}={{\left( x+y \right)}^{3}}-3{{\text{x}}^{2}}y-3\text{x}{{y}^{2}}+{{z}^{3}}={{\left( -z \right)}^{3}}-3\text{x}y\left( x+y \right)+{{z}^{3}}=3\text{x}y\Rightarrow P=\frac{3\text{xyz}}{-2\text{x}yz}=-\frac{3}{2}

II. Bài tập tự luyện

Câu  1: Rút gọn các biểu thức:

           a) A = 3−61−2−2+81+2\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}

           b) B = (1x−4−1x + 4x+4).x + 2xx\left( \frac{1}{\text{x}-4}-\frac{1}{\text{x + 4}\sqrt{\text{x}}+4} \right).\frac{\text{x + 2}\sqrt{\text{x}}}{\sqrt{\text{x}}}    ( với x > 0, x ≠\ne 4 ).

Câu  2: Rút gọn các biểu thức sau:

         a) A = (2+3+33+1).(2−3−33−1)\left( 2+\frac{3+\sqrt{3}}{\sqrt{3}+1} \right).\left( 2-\frac{3-\sqrt{3}}{\sqrt{3}-1} \right)

         b) B = (ba - ab - aab - b).(ab - ba)\left( \frac{\sqrt{\text{b}}}{\text{a - }\sqrt{\text{ab}}}\text{ - }\frac{\sqrt{\text{a}}}{\sqrt{\text{ab}}\text{ - b}} \right)\text{.}\left( \text{a}\sqrt{\text{b}}\text{ - b}\sqrt{\text{a}} \right)   ( với a > 0, b > 0, a ≠\ne b)

Câu  3: Cho biểu thức A = (aa−1−aa - a):a+1a - 1\left( \frac{\sqrt{\text{a}}}{\sqrt{\text{a}}-1}-\frac{\sqrt{\text{a}}}{\text{a - }\sqrt{\text{a}}} \right):\frac{\sqrt{\text{a}}+1}{\text{a - 1}}   với a > 0, a ≠\ne 1

    a) Rút gọn biểu thức A.

    b) Tìm các giá trị của a để A < 0.

Câu  4: a) Rút gọn biểu thức:   A = (3x+6x - 4+xx−2):x - 9x−3\left( \frac{3\sqrt{\text{x}}+6}{\text{x - 4}}+\frac{\sqrt{\text{x}}}{\sqrt{\text{x}}-2} \right):\frac{\text{x - 9}}{\sqrt{\text{x}}-3}    với x ≥ 0, x ≠ 4, x ≠ 9\text{x }\ge \text{ 0, x }\ne \text{ 4, x }\ne \text{ 9}.

Câu  5: Rút gọn các biểu thức:

         a) A = 38−50−(2−1)23\sqrt{8}-\sqrt{50}-\sqrt{{{\left( \sqrt{2}-1 \right)}^{2}}}

         b) B = 2x - 1.x2 - 2x + 14x2\frac{2}{\text{x - 1}}.\sqrt{\frac{{{\text{x}}^{\text{2}}}\text{ - 2x + 1}}{4{{\text{x}}^{\text{2}}}}}, với 0 < x < 1

Câu 6: 1) Rút gọn biểu thức:

              A = (1 − aa1 − a + a)(1 − a1 − a)2A\text{ }=\text{ }\left( \frac{1\text{ }-\text{ }a\sqrt{a}}{1\text{ }-\text{ }\sqrt{a}}\text{ }+\text{ }\sqrt{a} \right){{\left( \frac{1\text{ }-\text{ }\sqrt{a}}{1\text{ }-\text{ }a} \right)}^{2}} với a ≥ 0 và a ≠ 1.

Câu 7: Tính gọn biểu thức:

          1) A = 20 - 45 + 318 + 72\sqrt{20}\text{ - }\sqrt{45}\text{ + 3}\sqrt{18}\text{ + }\sqrt{72}.

          2) B = (1 + a + aa + 1)(1 + a - a 1&ThinSpace;-&ThinSpace;&ThinSpace;a)\left( 1\text{ + }\frac{\text{a + }\sqrt{\text{a}}}{\sqrt{\text{a}}\text{ + 1}} \right)\left( \text{1 + }\frac{\text{a - }\sqrt{\text{a}}}{\text{ 1}\,\text{-}\,\,\sqrt{a}} \right) với a ≥ 0, a ≠ 1.

Câu 8: Cho biểu thức:  P = (aa - 1a - a - aa + 1a + a) : a +2a - 2\left( \frac{\text{a}\sqrt{\text{a}}\text{ - 1}}{\text{a - }\sqrt{\text{a}}}\text{ - }\frac{\text{a}\sqrt{\text{a}}\text{ + 1}}{\text{a + }\sqrt{\text{a}}} \right)\text{ : }\frac{\text{a +}\text{2}}{\text{a - 2}} với  a >  0, a  ¹ 1, a ¹ 2.

             1) Rút gọn P.

             2) Tìm giá trị nguyên của a để P có giá trị nguyên.

Câu 9: Cho biểu thức

P = x + 1x - 2 + 2xx + 2 + 2 + 5x4 - x\frac{\sqrt{\text{x}}\text{ + 1}}{\sqrt{\text{x}}\text{ - 2}}\text{ + }\frac{2\sqrt{\text{x}}}{\sqrt{\text{x}}\text{ + 2}}\text{ + }\frac{2\text{ + 5}\sqrt{\text{x}}}{\text{4 - x}} với x ≥ 0, x ≠ 4.

1) Rút gọn P.

2) Tìm x để P = 2.

Câu 10: Cho M = (xx - 1 - 1x - x) : (1x+1 + 2x - 1)\left( \frac{\sqrt{\text{x}}}{\sqrt{\text{x}}\text{ - 1}}\text{ - }\frac{1}{\text{x - }\sqrt{\text{x}}} \right)\text{ : }\left( \frac{1}{\sqrt{\text{x}}+1}\text{ + }\frac{2}{\text{x - 1}} \right)   với  x&gt;0,&ThinSpace;&ThinSpace;x≠1x&gt;0,\,\,x\ne 1.

a) Rút gọn M.

b) Tìm x sao cho M > 0.

Câu 11: Cho biểu thức:  K = xx - 1 - 2x - xx - x\frac{\text{x}}{\sqrt{\text{x}}\text{ - }1}\text{ - }\frac{2\text{x - }\sqrt{\text{x}}}{\text{x - }\sqrt{\text{x}}}   với x >0 và  x≠\ne 1

Rút gọn biểu thức K

Tìm giá trị của biểu thức K tại x = 4 + 23\sqrt{3}

Câu 12:  Rút gọn các biểu thức:

            1) 45+20−5\sqrt{45}+\sqrt{20}-\sqrt{5}.

               2)    x+xx+x−4x+2\frac{\text{x}+\sqrt{\text{x}}}{\sqrt{\text{x}}}+\frac{\text{x}-4}{\sqrt{\text{x}}+2}  với x > 0.

Câu 13: Cho các biểu thức A = 5+755+11+111+11&ThinSpace;&ThinSpace;&ThinSpace;,&ThinSpace;&ThinSpace;&ThinSpace;&ThinSpace;B=5:55+55\frac{5+7\sqrt{5}}{\sqrt{5}}+\frac{11+\sqrt{11}}{1+\sqrt{11}}\,\,\,,\,\,\,\,B=\sqrt{5}:\frac{5}{5+\sqrt{55}}

a) Rút gọn biểu thức A.

b) Chứng minh: A - B = 7.

Câu 14: Rút gọn các biểu thức :

a) A = 25 - 2 - 25 + 2\frac{2}{\sqrt{5}\text{ - 2}}\text{ - }\frac{2}{\sqrt{5}\text{ + 2}}

b) B = (x - 1x) : (x - 1x + 1 - xx + x)\left( \sqrt{\text{x}}\text{ - }\frac{1}{\sqrt{\text{x}}} \right)\text{ : }\left( \frac{\sqrt{\text{x}}\text{ - 1}}{\sqrt{\text{x}}}\text{ + }\frac{1\text{ - }\sqrt{\text{x}}}{\text{x + }\sqrt{\text{x}}} \right)  với  x&gt;0,&ThinSpace;&ThinSpace;x≠1.x&gt;0,\,\,x\ne 1.

Câu 15: Cho biểu thức:  P = (a2−12a)(a−aa+1−a+aa−1)\left( \frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}} \right)\left( \frac{a-\sqrt{a}}{\sqrt{a}+1}-\frac{a+\sqrt{a}}{\sqrt{a}-1} \right)          

1) Rút gọn biểu thức P.

2) Tìm a để P > - 2

Câu 16. Rút gọn:

 A = (1−5)⋅5+525.(1-\sqrt{5})\cdot \frac{\sqrt{5}+5}{2\sqrt{5}}.

 B = (1+x+x1+x)(1+x−x1−x)\left( 1+\frac{x+\sqrt{x}}{1+\sqrt{x}} \right)\left( 1+\frac{x-\sqrt{x}}{1-\sqrt{x}} \right)   với 0≤x≠10\le x\ne 1.

Câu 17. Cho biểu thức A = (xx−1−1x−x):(1x+1+2x−1)\left( \frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}} \right):\left( \frac{1}{\sqrt{x}+1}+\frac{2}{x-1} \right) 

            1) Rút gọn biểu thức A.

            2) Tính giá trị của A khi x=22+3x=2\sqrt{2}+3.

Câu 18: Cho biểu thức P = (1x + x−1x+1):xx + 2x+1\left( \frac{\text{1}}{\text{x + }\sqrt{\text{x}}}-\frac{1}{\sqrt{\text{x}}+1} \right):\frac{\sqrt{\text{x}}}{\text{x + 2}\sqrt{\text{x}}+1}   với x > 0.Rút gọn biểu thức P.

Tìm các giá trị của x để P > 12\frac{1}{2}.

Câu 19: Rút gọn các biểu thức sau:

            1) A =  1220−80+2345\frac{1}{2}\sqrt{20}-\sqrt{80}+\frac{2}{3}\sqrt{45}

            2) B = (2+5−55−1).(2−5+55+1)\left( 2+\frac{5-\sqrt{5}}{\sqrt{5}-1} \right).\left( 2-\frac{5+\sqrt{5}}{\sqrt{5}+1} \right)

Câu 20: Cho biểu thức A = (aa+1−aa + a):a−1a - 1\left( \frac{\sqrt{\text{a}}}{\sqrt{\text{a}}+1}-\frac{\sqrt{\text{a}}}{\text{a + }\sqrt{\text{a}}} \right):\frac{\sqrt{\text{a}}-1}{\text{a - 1}}    với a > 0, a ≠\ne 1.

            1) Rút gọn biểu thức A.

            2) Tìm các giá trị của a để A < 0.

Câu 21: Cho biểu thức           P = (1a−3+1a+3)(1−3a)\left( \frac{1}{\sqrt{a}-3}+\frac{1}{\sqrt{a}+3} \right)\left( 1-\frac{3}{\sqrt{a}} \right) với a > 0 và a ≠\ne 9.

            a) Rút  gọn biểu thức P

            b) Tìm các giá trị của a  để P > 12\frac{1}{2}.

Câu 22. Rút gọn biểu thức P = 9a−25a+4a3a2+2a\frac{9\sqrt{a}-\sqrt{25a}+\sqrt{4{{a}^{3}}}}{{{a}^{2}}+2a} với a&gt;0a&gt;0.

Câu 23: Tính:

a) A=20−318−45+72A=\sqrt{20}-3\sqrt{18}-\sqrt{45}+\sqrt{72}.

b) B=4+7+4−7B=\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}.

c) C=x+2x−1+x−2x−1C=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}} với x > 1

Câu 24: 1) Rút gọn biểu thức:            P = (7+3−2)(7−3+2)(\sqrt{7}+\sqrt{3}-2)(\sqrt{7}-\sqrt{3}+2).

Câu 25: Cho biểu thức A =(1−2aa+1):(1a+1−2aaa+a+a+1)\left( 1-\frac{2\sqrt{a}}{a+1} \right):\left( \frac{1}{\sqrt{a}+1}-\frac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}+a+1} \right). Rút gọn biểu thức A.

Câu 26: Rút gọn biểu thức:    P = (a−1+1)2+(a−1−1)2\sqrt{{{(\sqrt{a-1}+1)}^{2}}}+\sqrt{{{(\sqrt{a-1}-1)}^{2}}} với  a > 1

Câu 27: Cho biểu thức:          Q = (x2−12x)2(x+1x−1−x−1x+1){{\left( \frac{\sqrt{x}}{2}-\frac{1}{2\sqrt{x}} \right)}^{2}}\left( \frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1} \right). Tìm tất cả các giá trị của x để Q có nghĩa. Rút gọn Q.

Câu 28: Rút gọn A = x2+6x+9x+3\frac{\sqrt{{{x}^{2}}+6x+9}}{x+3}  với  x≠−3x\ne -3.

Câu 29: a) Tính (1+5)2+(1−5)2\sqrt{{{(1+\sqrt{5})}^{2}}}+\sqrt{{{(1-\sqrt{5})}^{2}}}.

b) Giải phương trình: x2 + 2x - 24 = 0.

Câu 30: Cho biểu thức: P = 2aa+3+a+1a−3+3+7a9−a\frac{2\sqrt{a}}{\sqrt{a}+3}+\frac{\sqrt{a}+1}{\sqrt{a}-3}+\frac{3+7\sqrt{a}}{9-a}  với a > 0, a ≠\ne 9.

a) Rút gọn.

b) Tìm a để P < 1.

Câu 31: Cho biểu thức: M = x2−xx+x+1−x2+xx−x+1+x+1\frac{{{x}^{2}}-\sqrt{x}}{x+\sqrt{x}+1}-\frac{{{x}^{2}}+\sqrt{x}}{x-\sqrt{x}+1}+x+1

Rút gọn biểu thức M với x≥0.x\ge 0.

Câu 32: Cho biểu thức: P = x2+xx−x+1+1−2x+xx\frac{{{x}^{2}}+\sqrt{x}}{x-\sqrt{x}+1}+1-\frac{2x+\sqrt{x}}{\sqrt{x}} với  x > 0.Rút gọi biểu thức P.

Câu 33: 

Tính: 48−275+108\sqrt{48} - 2\sqrt{75} + \sqrt{108}

Rút gọn biểu thức: P= (11−x−11+x).(1−1x)\left( \frac{1}{1 - \sqrt{x}} - \frac{1}{1 + \sqrt{x}} \right) . \left( 1 - \frac{1}{\sqrt{x}} \right) với x≠\ne 1 và x >0

ĐÁP ÁN

Câu 1: a) A = 3−61−2−2+81+2=3(1−2)1−2−2(1+2)1+2=3−2\text{ a) A = }\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}=\frac{\sqrt{3}\left( 1-\sqrt{2} \right)}{1-\sqrt{2}}-\frac{2\left( 1+\sqrt{2} \right)}{1+\sqrt{2}}=\sqrt{3}-2

b) B = (1x−4−1x + 4x+4).x + 2xx\text{b) B = }\left( \frac{1}{\text{x}-4}-\frac{1}{\text{x + 4}\sqrt{\text{x}}+4} \right).\frac{\text{x + 2}\sqrt{\text{x}}}{\sqrt{\text{x}}} = (1(x−2)(x+2)−1(x+2)2).x(x + 2)x\text{= }\left( \frac{1}{\left( \sqrt{\text{x}}-2 \right)\left( \sqrt{\text{x}}+2 \right)}-\frac{1}{{{(\sqrt{\text{x}}+2)}^{2}}} \right).\frac{\sqrt{\text{x}}\text{(}\sqrt{\text{x}}\text{ + 2)}}{\sqrt{\text{x}}}

       =1x−2−1x+2=(x+2)−(x−2)x - 4=4x - 4\text{=}\frac{1}{\sqrt{\text{x}}-2}-\frac{1}{\sqrt{\text{x}}+2}=\frac{\left( \sqrt{\text{x}}+2 \right)-\left( \sqrt{\text{x}}-2 \right)}{\text{x - 4}}=\frac{4}{\text{x - 4}}

Câu 2:

  a) A = (2+3+33+1).(2−3−33−1)=(2+3(3+1)3+1)(2−3(3−1)3−1)=(2+3)(2−3)=1.\text{ a) A = }\left( 2+\frac{3+\sqrt{3}}{\sqrt{3}+1} \right).\left( 2-\frac{3-\sqrt{3}}{\sqrt{3}-1} \right)=\left( 2+\frac{\sqrt{3}\left( \sqrt{3}+1 \right)}{\sqrt{3}+1} \right)\left( 2-\frac{\sqrt{3}\left( \sqrt{3}-1 \right)}{\sqrt{3}-1} \right)=\left( 2+\sqrt{3} \right)\left( 2-\sqrt{3} \right)=1. 

b) (ba - ab - aab - b).(ab - ba)=(ba(a−b) - ab(a−b)).ab(a - b)\text{b) }\left( \frac{\sqrt{\text{b}}}{\text{a - }\sqrt{\text{ab}}}\text{ - }\frac{\sqrt{\text{a}}}{\sqrt{\text{ab}}\text{ - b}} \right)\text{.}\left( \text{a}\sqrt{\text{b}}\text{ - b}\sqrt{\text{a}} \right)=\left( \frac{\sqrt{\text{b}}}{\sqrt{\text{a}}\left( \sqrt{\text{a}}-\sqrt{\text{b}} \right)}\text{ - }\frac{\sqrt{\text{a}}}{\sqrt{\text{b}}\left( \sqrt{\text{a}}-\sqrt{\text{b}} \right)} \right)\text{.}\sqrt{\text{ab}}\left( \sqrt{\text{a}}\text{ - }\sqrt{\text{b}} \right)

=b.aba−a.abb=b - a.=\frac{\sqrt{\text{b}}.\sqrt{\text{ab}}}{\sqrt{\text{a}}}-\frac{\sqrt{\text{a}}.\sqrt{\text{ab}}}{\sqrt{\text{b}}}=\text{b - a}\text{.}

Câu 3:

  a) A =(aa−1−aa(a - 1)):a+1(a - 1)(a+1)=(aa−1−1(a - 1)).(a−1)=a−1\mathbf{ }\text{ a) A =}\left( \frac{\sqrt{\text{a}}}{\sqrt{\text{a}}-1}-\frac{\sqrt{\text{a}}}{\sqrt{\text{a}}\text{(}\sqrt{\text{a}}\text{ - 1)}} \right):\frac{\sqrt{\text{a}}+1}{(\sqrt{\text{a}}\text{ - 1)(}\sqrt{\text{a}}+1)}=\left( \frac{\sqrt{\text{a}}}{\sqrt{\text{a}}-1}-\frac{\text{1}}{\text{(}\sqrt{\text{a}}\text{ - 1)}} \right).\left( \sqrt{\text{a}}-1 \right)=\sqrt{\text{a}}-1

   b) A < 0

Câu  4: a) A =(3x+6x - 4+xx−2):x - 9x−3\left( \frac{3\sqrt{\text{x}}+6}{\text{x - 4}}+\frac{\sqrt{\text{x}}}{\sqrt{\text{x}}-2} \right):\frac{\text{x - 9}}{\sqrt{\text{x}}-3}

=(3(x+2)(x−2)(x+2)+xx−2):(x−3)(x+3)x−3=\left( \frac{3(\sqrt{\text{x}}+2)}{\left( \sqrt{\text{x}}-2 \right)\left( \sqrt{\text{x}}+2 \right)}+\frac{\sqrt{\text{x}}}{\sqrt{\text{x}}-2} \right):\frac{\left( \sqrt{\text{x}}-3 \right)\left( \sqrt{\text{x}}+3 \right)}{\sqrt{\text{x}}-3}

=(3+xx−2).1x+3=1x−2=\left( \frac{3+\sqrt{\text{x}}}{\sqrt{\text{x}}-2} \right).\frac{1}{\sqrt{\text{x}}+3}=\frac{1}{\sqrt{\text{x}}-2}, với x ≥ 0,&ThinSpace;&ThinSpace;x ≠ 4,&ThinSpace;&ThinSpace;x ≠ 9\text{x }\ge \text{ 0,}\,\,\text{x }\ne \text{ 4,}\,\,\text{x }\ne \text{ 9}.

Câu 5:

A=38−50−(2−1)2=62−52−(2−1)=1A=3\sqrt{8}-\sqrt{50}-\sqrt{{{\left( \sqrt{2}-1 \right)}^{2}}}=6\sqrt{2}-5\sqrt{2}-\left( \sqrt{2}-1 \right)=1

b) B = 2x - 1.x2 - 2x + 14x2=2x - 1(x - 1)222x2=2x - 1.∣x - 1∣2∣x∣\text{B = }\frac{2}{\text{x - 1}}.\sqrt{\frac{{{\text{x}}^{\text{2}}}\text{ - 2x + 1}}{4{{\text{x}}^{\text{2}}}}}=\frac{2}{\text{x - 1}}\sqrt{\frac{{{\left( \text{x - 1} \right)}^{2}}}{{{2}^{2}}{{\text{x}}^{\text{2}}}}}=\frac{2}{\text{x - 1}}.\frac{\left| \text{x - 1} \right|}{2\left| \text{x} \right|}

Vì 0 < x < 1 nên ⇒B = - 2(x - 1)2x(x - 1)=−1x\Rightarrow \text{B = }\frac{\text{- 2}\left( \text{x - 1} \right)}{\text{2x}\left( \text{x - 1} \right)}=-\frac{1}{\text{x}}.

Câu 6:  1) Rút gọn

A = [(1 - a) (1 + a + a)1 - a + a][1 - a(1 - a) (1 + a)]2\left[ \frac{\left( \text{1 - }\sqrt{\text{a}} \right)\text{ }\left( \text{1 + }\sqrt{\text{a}}\text{ + a} \right)}{\text{1 - }\sqrt{\text{a}}}\text{ + }\sqrt{\text{a}} \right]{{\left[ \frac{\text{1 - }\sqrt{\text{a}}}{\left( \text{1 - }\sqrt{\text{a}} \right)\text{ }\left( \text{1 + }\sqrt{\text{a}} \right)} \right]}^{2}}

  = (1 + 2a + a).1(1 + a)2 = (1 + a)2.1(1 +a)2 = 1.\left( \text{1 + 2}\sqrt{\text{a}}\text{ + a} \right).\frac{1}{{{\left( 1\text{ + }\sqrt{\text{a}} \right)}^{2}}}\text{ = }{{\left( 1\text{ + }\sqrt{\text{a}} \right)}^{2}}.\frac{1}{{{\left( 1\text{ +}\sqrt{\text{a}} \right)}^{2}}}\text{ = 1}\text{.}

Câu 7: Rút gọn biểu thức

1)   A = 20 - 45 + 318 + 72\sqrt{20}\text{ - }\sqrt{45}\text{ + 3}\sqrt{18}\text{ + }\sqrt{72} = 5 . 4 - 9 . 5 + 39 . 2 + 36 . 2\sqrt{5\text{ }\text{. 4}}\text{ - }\sqrt{\text{9 }\text{. 5}}\text{ + 3}\sqrt{9\text{ }\text{. 2}}\text{ + }\sqrt{\text{36 }\text{. 2}}

 = 25 - 35 + 92 + 622\sqrt{5}\text{ - 3}\sqrt{5}\text{ + 9}\sqrt{2}\text{ + 6}\sqrt{2} = 152 - 5\sqrt{2}\text{ - }\sqrt{5}

2) B = (1 + a + aa + 1)(1 + a - a1 - a)\left( 1\text{ + }\frac{\text{a + }\sqrt{\text{a}}}{\sqrt{\text{a}}\text{ + 1}} \right)\left( \text{1 + }\frac{\text{a - }\sqrt{\text{a}}}{1\text{ - }\sqrt{\text{a}}} \right) với a ≥ 0,   a ≠ 1

= (1 + a (a + 1)a + 1)(1 - a (a - 1)a - 1)\left( 1\text{ + }\frac{\sqrt{\text{a}}\text{ (}\sqrt{\text{a}}\text{ + 1)}}{\sqrt{\text{a}}\text{ + 1}} \right)\left( \text{1 - }\frac{\sqrt{\text{a}}\text{ (}\sqrt{\text{a}}\text{ - 1)}}{\sqrt{\text{a}}\text{ - 1}} \right) = (1 + a\sqrt{\text{a}}) (1 - a\sqrt{\text{a}}) = 1 - a

Câu 8:

1) Điều kiện:  a ≥ 0,  a ≠ 1, a ≠ 2

Ta có: P = [(a - 1) (a + a + 1)a (a - 1) - (a + 1) (a - a + 1)a (a + 1)] : a + 2a - 2\text{P = }\left[ \frac{\left( \sqrt{\text{a}}\text{ - 1} \right)\text{ }\left( \text{a + }\sqrt{\text{a}}\text{ + 1} \right)}{\sqrt{\text{a}}\text{ }\left( \sqrt{\text{a}}\text{ - 1} \right)}\text{ - }\frac{\left( \sqrt{\text{a}}\text{ + 1} \right)\text{ }\left( \text{a - }\sqrt{\text{a}}\text{ + 1} \right)}{\sqrt{\text{a}}\text{ }\left( \sqrt{\text{a}}\text{ + 1} \right)} \right]\text{ : }\frac{\text{a + 2}}{\text{a - 2}}

 = a + a + 1 - a + a - 1a : a + 2a - 2\text{ = }\frac{\text{a + }\sqrt{\text{a}}\text{ + 1 - a + }\sqrt{\text{a}}\text{ - 1}}{\sqrt{\text{a}}}\text{ : }\frac{\text{a + 2}}{\text{a - 2}}  = 2 (a - 2)a + 2\text{= }\frac{2\text{ (a - 2)}}{\text{a + 2}}

Câu 9: 1) Ta có : P = x + 1x - 2 + 2xx +2 - 2 + 5xx - 4\text{P = }\frac{\sqrt{\text{x}}\text{ + 1}}{\sqrt{\text{x}}\text{ - 2}}\text{ + }\frac{2\sqrt{\text{x}}}{\sqrt{\text{x}}\text{ +}\text{2}}\text{ - }\frac{2\text{ + 5}\sqrt{\text{x}}}{\text{x - 4}}

P = (x+1) (x +2) + 2x (x - 2) - 2 - 5x(x - 2) (x + 2)\frac{(\sqrt{\text{x}}\text{+1) (}\sqrt{\text{x}}\text{ +}\text{2) + 2}\sqrt{\text{x}}\text{ (}\sqrt{\text{x}}\text{ - 2) - 2 - 5}\sqrt{\text{x}}}{(\sqrt{\text{x}}\text{ - 2) (}\sqrt{\text{x}}\text{ + 2)}} =

  =   x + 3x +2 + 2x - 4x - 2 - 5x(x +2) (x - 2)\frac{\text{x + 3}\sqrt{\text{x}}\text{ +}\text{2 + 2x - 4}\sqrt{\text{x}}\text{ - 2 - 5}\sqrt{\text{x}}}{(\sqrt{\text{x}}\text{ +}\text{2) (}\sqrt{\text{x}}\text{ - 2)}}

  = 3x - 6x(x + 2) (x - 2) = 3x (x−2)(x + 2) (x - 2) = 3xx +2\frac{\text{3x - 6}\sqrt{\text{x}}}{(\sqrt{\text{x}}\text{ + 2) (}\sqrt{\text{x}}\text{ - 2)}}\text{ = }\frac{3\sqrt{\text{x}}\text{ (}\sqrt{\text{x}}-2)}{(\sqrt{\text{x}}\text{ + 2) (}\sqrt{\text{x}}\text{ - 2)}}\text{ = }\frac{3\sqrt{\text{x}}}{\sqrt{\text{x}}\text{ +}\text{2}}

Câu 10: a) M = (xx - 1 - 1x - x) : (1x + 1 + 2x - 1)\left( \frac{\sqrt{\text{x}}}{\sqrt{\text{x}}\text{ - 1}}\text{ - }\frac{1}{\text{x - }\sqrt{\text{x}}} \right)\text{ : }\left( \frac{1}{\sqrt{\text{x}}\text{ + 1}}\text{ + }\frac{2}{\text{x - 1}} \right)

 = [xx - 1 - 1x (x - 1)] : [x - 1(x - 1) (x + 1) + 2(x - 1) (x +1)]\left[ \frac{\sqrt{\text{x}}}{\sqrt{\text{x}}\text{ - 1}}\text{ - }\frac{1}{\sqrt{\text{x}}\text{ (}\sqrt{\text{x}}\text{ - 1)}} \right]\text{ : }\left[ \frac{\sqrt{\text{x}}\text{ - 1}}{\left( \sqrt{\text{x}}\text{ - 1} \right)\text{ }\left( \sqrt{\text{x}}\text{ + 1} \right)}\text{ + }\frac{2}{\left( \sqrt{\text{x}}\text{ - 1} \right)\text{ }\left( \sqrt{\text{x}}\text{ +}\text{1} \right)} \right]

 = x - 1x (x - 1) : x + 1(x - 1) (x +1) = x - 1x(x - 1) . (x - 1) (x + 1)x + 1\frac{\text{x - 1}}{\sqrt{\text{x}}\text{ }\left( \sqrt{\text{x}}\text{ - 1} \right)}\text{ : }\frac{\sqrt{\text{x}}\text{ + 1}}{\left( \sqrt{\text{x}}\text{ - 1} \right)\text{ }\left( \sqrt{\text{x}}\text{ +}\text{1} \right)}\text{ = }\frac{\text{x - 1}}{\sqrt{\text{x}}\left( \sqrt{\text{x}}\text{ - 1} \right)}\text{ }\text{. }\frac{\left( \sqrt{\text{x}}\text{ - 1} \right)\text{ }\left( \sqrt{\text{x}}\text{ + 1} \right)}{\sqrt{\text{x}}\text{ + 1}}

 = x - 1x\frac{\text{x - 1}}{\sqrt{\text{x}}}.

Câu 11:

1)  K = xx - 1-x(2x - 1)x(x - 1)\frac{\text{x}}{\sqrt{\text{x}}\text{ - 1}}\text{-}\frac{\sqrt{\text{x}}\text{(2}\sqrt{\text{x}}\text{ - 1)}}{\sqrt{\text{x}}\text{(}\sqrt{\text{x}}\text{ - 1)}}  = x - 2x + 1x - 1 = x - 1\frac{\text{x - 2}\sqrt{\text{x}}\text{ + 1}}{\sqrt{\text{x}}\text{ - 1}}\text{ = }\sqrt{\text{x}}\text{ - 1}

2) Khi  x = 4 + 23\sqrt{3}, ta có: K = 4+23\sqrt{4+2\sqrt{3}}- 1 =  (3+1)2−1=3+1−1=3\sqrt{{{\left( \sqrt{3}+1 \right)}^{2}}}-1=\sqrt{3}+1-1=\sqrt{3}

Câu 12: Rút gọn biểu thức:

1) 45+20−5\sqrt{45}+\sqrt{20}-\sqrt{5} = 32.5+22.5−5\sqrt{{{3}^{2}}.5}+\sqrt{{{2}^{2}}.5}-\sqrt{5}

 = 35+25−53\sqrt{5}+2\sqrt{5}-\sqrt{5}  =  45\sqrt{5}

2)  x+xx+x−4x+2\frac{x+\sqrt{x}}{\sqrt{x}}+\frac{x-4}{\sqrt{x}+2}= x(x+1)x+(x+2)(x−2)x+2\frac{\sqrt{x}(\sqrt{x}+1)}{\sqrt{x}}+\frac{(\sqrt{x}+2)(\sqrt{x}-2)}{\sqrt{x}+2}

= x+1+x−2\sqrt{x}+1+\sqrt{x}-2 =  2x−1\sqrt{x}-1

Câu 13: a)  A = 5(5+7)5+11(11+1)1+11=5+7+11.\frac{\sqrt{5}(\sqrt{5}+7)}{\sqrt{5}}+\frac{\sqrt{11}(\sqrt{11}+1)}{1+\sqrt{11}}=\sqrt{5}+7+\sqrt{11}.

b)  B = 5.5(5+11)5=5+11\sqrt{5}.\frac{\sqrt{5}(\sqrt{5}+\sqrt{11})}{5}=\sqrt{5}+\sqrt{11}.

Vậy  A−B=5+7+11−5−11=7A-B=\sqrt{5}+7+\sqrt{11}-\sqrt{5}-\sqrt{11}=7

Câu 14: a) A = 2(5 +2) - 2(5 - 2)(5 - 2) (5 +2) = 25 +4 - 25 + 4(5)2 - 22 = 85 - 4 = 8\text{A = }\frac{2(\sqrt{5}\text{ +}2)\text{ - 2(}\sqrt{5}\text{ - 2)}}{\left( \sqrt{5}\text{ - 2} \right)\text{ }\left( \sqrt{5}\text{ +}\text{2} \right)}\text{ = }\frac{2\sqrt{5}\text{ +}\text{4 - 2}\sqrt{5}\text{ + 4}}{{{\left( \sqrt{5} \right)}^{2}}\text{ - }{{\text{2}}^{2}}}\text{ = }\frac{8}{5\text{ - 4}}\text{ = 8}.

b) Ta có:

B = x - 1x:(x - 1)(x + 1) +1 - xx(x +1) = x - 1x⋅x(x +1)x - 1 + 1 - x &ThinSpace;&ThinSpace;= (x - 1)(x +&ThinSpace;1)x (x - 1)=(x +&ThinSpace;1)2x \text{B = }\frac{\text{x - 1}}{\sqrt{\text{x}}}\text{:}\frac{\left( \sqrt{\text{x}}\text{ - 1} \right)\left( \sqrt{\text{x}}\text{ + 1} \right)\text{ +}\text{1 - }\sqrt{\text{x}}}{\sqrt{\text{x}}\left( \sqrt{\text{x}}\text{ +}\text{1} \right)}\text{ = }\frac{\text{x - 1}}{\sqrt{\text{x}}}\cdot \frac{\sqrt{\text{x}}\left( \sqrt{\text{x}}\text{ +}\text{1} \right)}{\text{x - 1 + 1 - }\sqrt{\text{x}}}\text{ }\,\,\text{= }\frac{\left( \text{x - 1} \right)\left( \sqrt{\text{x}}\text{ +}\,\text{1} \right)}{\sqrt{\text{x}}\text{ }\left( \sqrt{\text{x}}\text{ - 1} \right)}=\frac{{{\left( \sqrt{\text{x}}\text{ +}\,\text{1} \right)}^{2}}}{\sqrt{\text{x}}\text{ }}

Câu 15.

 1) A = (1−5)⋅5(1+5)25=(1−5)⋅(1+5)2=1−52=−2(1-\sqrt{5})\cdot \frac{\sqrt{5}(1+\sqrt{5})}{2\sqrt{5}}=(1-\sqrt{5})\cdot \frac{(1+\sqrt{5})}{2}=\frac{1-5}{2}=-2.

 2) B = (1+x(x+1)1+x)(1+x(x−1)1−x)=(1+x)(1−x)=1−x\left( 1+\frac{\sqrt{x}\left( \sqrt{x}+1 \right)}{1+\sqrt{x}} \right)\left( 1+\frac{\sqrt{x}\left( \sqrt{x}-1 \right)}{1-\sqrt{x}} \right)=\left( 1+\sqrt{x} \right)\left( 1-\sqrt{x} \right)=1-x.

Câu 16.

 1) Ta có A = (x−1x(x−1)):(x+1x−1)\left( \frac{x-1}{\sqrt{x}\left( \sqrt{x}-1 \right)} \right):\left( \frac{\sqrt{x}+1}{x-1} \right) = x+1x.x−1x+1=x−1x\frac{\sqrt{x}+1}{\sqrt{x}}.\frac{x-1}{\sqrt{x}+1}=\frac{x-1}{\sqrt{x}}.

Câu 17:

 1)  P = (1x + x−1x+1):xx + 2x+1\text{ P = }\left( \frac{\text{1}}{\text{x + }\sqrt{\text{x}}}-\frac{1}{\sqrt{\text{x}}+1} \right):\frac{\sqrt{\text{x}}}{\text{x + 2}\sqrt{\text{x}}+1} =(1x(x+1)−xx(x+1)).(x+1)2x=\left( \frac{1}{\sqrt{\text{x}}\left( \sqrt{\text{x}}+1 \right)}-\frac{\sqrt{\text{x}}}{\sqrt{\text{x}}\left( \sqrt{\text{x}}+1 \right)} \right).\frac{{{\left( \sqrt{\text{x}}+1 \right)}^{2}}}{\sqrt{\text{x}}}

 =1−xx(x+1).(x+1)2x=(1−x)(x+1)x.x=1 - x x=\frac{1-\sqrt{\text{x}}}{\sqrt{\text{x}}\left( \sqrt{\text{x}}+1 \right)}.\frac{{{\left( \sqrt{\text{x}}+1 \right)}^{2}}}{\sqrt{\text{x}}}=\frac{\left( 1-\sqrt{\text{x}} \right)\left( \sqrt{\text{x}}+1 \right)}{\sqrt{\text{x}}.\sqrt{\text{x}}}=\frac{\text{1 - x }}{\text{x}}.

 2) Với x > 0 thì 1 - xx&gt;12⇔2(1 - x)&gt;x⇔-3x+2&gt;0⇔x&lt;23\frac{\text{1 - x}}{\text{x}}&gt;\frac{1}{2}\Leftrightarrow 2\left( \text{1 - x} \right)&gt;\text{x}\Leftrightarrow \text{-3x+}2&gt;0\Leftrightarrow x&lt;\frac{2}{3}

Câu 18:

 1) A = 124.5−16.5+239.5\frac{1}{2}\sqrt{4.5}-\sqrt{16.5}+\frac{2}{3}\sqrt{9.5}  = 5−45+25\sqrt{5}-4\sqrt{5}+2\sqrt{5} = −5-\sqrt{5}.

2) B = (2+5−55−1).(2−5+55+1)\text{B = }\left( 2+\frac{5-\sqrt{5}}{\sqrt{5}-1} \right).\left( 2-\frac{5+\sqrt{5}}{\sqrt{5}+1} \right) 

       =(2+5(5−1)5−1)(2−5(5+1)5+1)=(2+5)(2−5)=−1=\left( 2+\frac{\sqrt{5}\left( \sqrt{5}-1 \right)}{\sqrt{5}-1} \right)\left( 2-\frac{\sqrt{5}\left( \sqrt{5}+1 \right)}{\sqrt{5}+1} \right)=\left( 2+\sqrt{5} \right)\left( 2-\sqrt{5} \right)=-1

Câu 19.

1) A =(aa+1−aa(a + 1)):a−1(a - 1)(a+1)\text{ A =}\left( \frac{\sqrt{\text{a}}}{\sqrt{\text{a}}+1}-\frac{\sqrt{\text{a}}}{\sqrt{\text{a}}\text{(}\sqrt{\text{a}}\text{ + 1)}} \right):\frac{\sqrt{\text{a}}-1}{(\sqrt{\text{a}}\text{ - 1)(}\sqrt{\text{a}}+1)}   =(aa+1−1a + 1).(a+1)=a−1\text{ }=\left( \frac{\sqrt{\text{a}}}{\sqrt{\text{a}}+1}-\frac{\text{1}}{\sqrt{\text{a}}\text{ + 1}} \right).\left( \sqrt{\text{a}}+1 \right)=\sqrt{\text{a}}-1

 2) A < 0

Câu 20:

a)  P = (1a−3+1a+3).(1−3a)=a+3+a−3(a−3)(a+3).a−3a\left( \frac{1}{\sqrt{a}-3}+\frac{1}{\sqrt{a}+3} \right).\left( 1-\frac{3}{\sqrt{a}} \right)=\frac{\sqrt{a}+3+\sqrt{a}-3}{\left( \sqrt{a}-3 \right)\left( \sqrt{a}+3 \right)}.\frac{\sqrt{a}-3}{\sqrt{a}}.

  = 2a.(a−3)(a−3)(a+3).a=2a+3\frac{2\sqrt{a}.(\sqrt{a}-3)}{(\sqrt{a}-3)(\sqrt{a}+3).\sqrt{a}}=\frac{2}{\sqrt{a}+3}. Vậy P = 2a+3\frac{2}{\sqrt{a}+3}.

b) Ta có: 2a+3\frac{2}{\sqrt{a}+3} > 12\frac{1}{2} ⇔\Leftrightarrow a\sqrt{a} + 3 < 4  ⇔\Leftrightarrow a\sqrt{a} < 1 

 Vậy P > 12\frac{1}{2} khi và chỉ khi 0 < a < 1.

Câu 21. 1) Ta có 9a−25a+4a3=9a−5a+2aa9\sqrt{a}-\sqrt{25a}+\sqrt{4{{a}^{3}}}=9\sqrt{a}-5\sqrt{a}+2a\sqrt{a} =2a(a+2)=2\sqrt{a}(a+2) và a2+2a=a(a+2){{a}^{2}}+2a=a(a+2)

nên P = 2a(a+2)a(a+2)=2a\frac{2\sqrt{a}\left( a+2 \right)}{a\left( a+2 \right)}=\frac{2}{\sqrt{a}} .

Câu 22: Tính

a) A = 20−318−45+72=4.5−39.2−9.5+36.2\sqrt{20}-3\sqrt{18}-\sqrt{45}+\sqrt{72}=\sqrt{4.5}-3\sqrt{9.2}-\sqrt{9.5}+\sqrt{36.2} =

=  25−92−35+62=−32−52\sqrt{5}-9\sqrt{2}-3\sqrt{5}+6\sqrt{2}=-3\sqrt{2}-\sqrt{5}.

b) B = 4+7+4−7\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}

2B=8+27+8−27=(7+1)2+(7−1)2=7+1+∣7−1∣\sqrt{2}B=\sqrt[{}]{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}=\sqrt{{{(\sqrt{7}+1)}^{2}}}+\sqrt{{{(\sqrt{7}-1)}^{2}}}=\sqrt{7}+1+\left| \sqrt{7}-1 \right|         

2B=27⇔B=14\sqrt{2}B=2\sqrt{7}\Leftrightarrow B=\sqrt{14}

c) C = x+2x−1+x−2x−1\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}} với x > 1

C = (x−1+1)2+(x−1−1)2=x−1+1+∣x−1−1∣\sqrt{{{(\sqrt{x-1}+1)}^{2}}}+\sqrt{{{(\sqrt{x-1}-1)}^{2}}}=\sqrt{x-1}+1+\left| \sqrt{x-1}-1 \right|

        +) Nếu  x > 2 thì C = x−1+1+x−1−1=2x−1\sqrt{x-1}+1+\sqrt{x-1}-1=2\sqrt{x-1}

+) Nếu  x < 2, thì C = x−1+1+1−x−1=2\sqrt{x-1}+1+1-\sqrt{x-1}=2.

Câu 23: 1) P = (7+3−2)(7−3+2)= &NegativeThinSpace;&NegativeThinSpace;[&NegativeThinSpace;&NegativeThinSpace; 7+(3−2) &NegativeThinSpace;&NegativeThinSpace;]&NegativeThinSpace;&NegativeThinSpace; &ThinSpace; &NegativeThinSpace;&NegativeThinSpace;[&NegativeThinSpace;&NegativeThinSpace; 7−(3−2) &NegativeThinSpace;&NegativeThinSpace;]&NegativeThinSpace;&NegativeThinSpace; (\sqrt{7}+\sqrt{3}-2)(\sqrt{7}-\sqrt{3}+2)=\text{ }\!\![\!\!\text{ }\sqrt{7}+(\sqrt{3}-2)\text{ }\!\!]\!\!\text{ }\,\text{ }\!\![\!\!\text{ }\sqrt{7}-(\sqrt{3}-2)\text{ }\!\!]\!\!\text{ }

=  (7)2−(3−2))2=&ThinSpace;7−(3−43+4)=43{{(\sqrt{7})}^{2}}-(\sqrt{3}-2){{)}^{2}}=\,7-(3-4\sqrt{3}+4)=4\sqrt{3}.

Câu 24: a) A = (a+1−2aa+1):[1a+1−2aa(a+1)+(a+1)]\left( \frac{a+1-2\sqrt{a}}{a+1} \right):\left[ \frac{1}{\sqrt{a}+1}-\frac{2\sqrt{a}}{\sqrt{a}(a+1)+(a+1)} \right]

= (a−1)2a+1:[1a+1−2a(a+1)(a+1)]=(a−1)2a+1:(a−1)2(a+1)(a+1)\frac{{{(\sqrt{a}-1)}^{2}}}{a+1}:\left[ \frac{1}{\sqrt{a}+1}-\frac{2\sqrt{a}}{(a+1)(\sqrt{a}+1)} \right]=\frac{{{(\sqrt{a}-1)}^{2}}}{a+1}:\frac{{{(\sqrt{a}-1)}^{2}}}{(\sqrt{a}+1)(a+1)}.

= (a−1)2a+1.(a+1)(a+1)(a−1)2=a+1\frac{{{(\sqrt{a}-1)}^{2}}}{a+1}.\frac{(a+1)(\sqrt{a}+1)}{{{(\sqrt{a}-1)}^{2}}}=\sqrt{a}+1.

b) a = 2011 - 2 2010=(2010−1)2⇒a=2010−1\sqrt{2010}={{(\sqrt{2010}-1)}^{2}}\Rightarrow \sqrt{a}=\sqrt{2010}-1

Vậy A = 2010\sqrt{2010}.

Câu 25: P = ∣a−1+1∣+∣a−1−1∣\left| \sqrt{a}-1+1 \right|+\left| \sqrt{a}-1-1 \right|

Nếu a> 2 => a−1−1≥0⇒P=2a−1\sqrt{a}-1-1\ge 0\Rightarrow P=2\sqrt{a}-1

Nếu 1< a < 2 => a−1−1\sqrt{a}-1-1 < 0 => P = 2

Câu 26: ĐKXĐ: x > 0; x ≠\ne 1.

1) Q = (x−1)24x.(x+1)2−(x−1)2x−1=(x−1)2.4x4x.(x−1)=x−1x\frac{{{(x-1)}^{2}}}{4x}.\frac{{{(\sqrt{x}+1)}^{2}}-{{(\sqrt{x}-1)}^{2}}}{x-1}=\frac{{{(x-1)}^{2}}.4\sqrt{x}}{4x.(x-1)}=\frac{x-1}{\sqrt{x}}.

2) Q = - 3x−3\sqrt{x}-3 => 4x + 3x\sqrt{x} - 1 = 0 (thỏa mãn)

Câu 27: A = (x+3)2x+3=∣x+3∣x+3\frac{\sqrt{{{(x+3)}^{2}}}}{x+3}=\frac{\left| x+3 \right|}{x+3} = 

Câu 28: a) P = 2aa+3+a+1a−3+−7a−3(a−3)(a+3)\frac{2\sqrt{a}}{\sqrt{a}+3}+\frac{\sqrt{a}+1}{\sqrt{a}-3}+\frac{-7\sqrt{a}-3}{(\sqrt{a}-3)(\sqrt{a}+3)}

= 2a(a−3)+(a+1)(a+3)−7a−3(a−3)(a+3)=2a−6a+a+4a+3−7a−3(a−3)(a+3)\frac{2\sqrt{a}(\sqrt{a}-3)+(\sqrt{a}+1)(\sqrt{a}+3)-7\sqrt{a}-3}{(\sqrt{a}-3)(\sqrt{a}+3)}=\frac{2a-6\sqrt{a}+a+4\sqrt{a}+3-7\sqrt{a}-3}{(\sqrt{a}-3)(\sqrt{a}+3)}

= 3a−9a(a−3)(a+3)=3a(a−3)(a−3)(a+3)=3aa+3\frac{3a-9\sqrt{a}}{(\sqrt{a}-3)(\sqrt{a}+3)}=\frac{3\sqrt{a}(\sqrt{a}-3)}{(\sqrt{a}-3)(\sqrt{a}+3)}=\frac{3\sqrt{a}}{\sqrt{a}+3}

Vậy  P = 3aa+3\frac{3\sqrt{a}}{\sqrt{a}+3}.

b)  P < 1 ⇔\Leftrightarrow 3aa+3&lt;1⇔3a&lt;a+3⇔a&lt;32⇔0≤a&lt;94\frac{3\sqrt{a}}{\sqrt{a}+3}&lt;1\Leftrightarrow 3\sqrt{a}&lt;\sqrt{a}+3\Leftrightarrow \sqrt{a}&lt;\frac{3}{2}\Leftrightarrow 0\le a&lt;\frac{9}{4}.

Câu 29: M = x(x3−1)x+x+1−x(x3+1)x−x+1\frac{\sqrt{x}(\sqrt{{{x}^{3}}}-1)}{x+\sqrt{x}+1}-\frac{\sqrt{x}(\sqrt{{{x}^{3}}}+1)}{x-\sqrt{x}+1}+ x + 1

= x(x−1)(x+x+1)x+x+1−x(x+1)(x−x+1)x−x+1+x+1\frac{\sqrt{x}(\sqrt{x}-1)(x+\sqrt{x}+1)}{x+\sqrt{x}+1}-\frac{\sqrt{x}(\sqrt{x}+1)(x-\sqrt{x}+1)}{x-\sqrt{x}+1}+x+1

= x - x\sqrt{x}- x - x\sqrt{x} + x + 1 = x - 2x\sqrt{x} + 1 = (x\sqrt{x} - 1)2

Câu 30:

a) Ta có x2 + x=x(x3+1)=x(x+1)(x−x+1)\sqrt{x}=\sqrt{x}(\sqrt{{{x}^{3}}}+1)=\sqrt{x}(\sqrt{x}+1)(x-\sqrt{x}+1)

nên  P = x(x+1)(x−x+1)x−x+1+1−x(2x+1)x\frac{\sqrt{x}(\sqrt{x}+1)(x-\sqrt{x}+1)}{x-\sqrt{x}+1}+1-\frac{\sqrt{x}(2\sqrt{x}+1)}{\sqrt{x}}

= x(x+1)+1−2x−1=x−x\sqrt{x}(\sqrt{x}+1)+1-2\sqrt{x}-1=x-\sqrt{x}. Vậy P = x−xx-\sqrt{x}.

b) P = 0 ⇔\Leftrightarrow   x - x\sqrt{x} = 0 ⇔\Leftrightarrow x\sqrt{x}(x\sqrt{x} - 1) = 0  ⇔\Leftrightarrow           x = 0 (loại) ; x = 1 (t/m)

Vậy x = 1 thì P = 0

Câu 31:

1) Tính: 48 - 275 + 108\sqrt{\text{48}}\text{ - 2}\sqrt{\text{75}}\text{ + }\sqrt{\text{108}}= 16 . 3 - 225 . 3 + 36 . 3\sqrt{\text{16 }\text{. 3}}\text{ - 2}\sqrt{\text{25 }\text{. 3}}\text{ + }\sqrt{\text{36 }\text{. 3}}

 =  43 - 103 + 63 = 0\text{4}\sqrt{\text{3}}\text{ - 10}\sqrt{\text{3}}\text{ + 6}\sqrt{\text{3}}\text{ = 0}

2) Rút gọn biểu thức:  P = (11 - x - 11 + x) . (1 - 1x)\left( \frac{\text{1}}{\text{1 - }\sqrt{\text{x}}}\text{ - }\frac{\text{1}}{\text{1 + }\sqrt{\text{x}}} \right)\text{ }.\text{ }\left( \text{1 - }\frac{\text{1}}{\sqrt{\text{x}}} \right)

= (1 + x - 1 +x1- x)(x - 1x)\left( \frac{\text{1 + }\sqrt{\text{x}}\text{ - 1 +}\sqrt{\text{x}}}{\text{1- x}} \right)\left( \frac{\sqrt{\text{x}}\text{ - 1}}{\sqrt{\text{x}}} \right) =2x1- x . x - 1x\frac{\text{2}\sqrt{\text{x}}}{\text{1- x}}\text{ }\text{. }\frac{\sqrt{\text{x}}\text{ - 1}}{\sqrt{\text{x}}}  = - 21 + x\frac{\text{- 2}}{\text{1 + }\sqrt{\text{x}}}

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