Đáp án+Giải thích các bước giải:
Bài 2:
`a)`
`A=(a-b)^2-c^2`
`=(a-b-c).(a-b+c)`
`b)`
`B=(m+n)^2-(m-n)^2`
`=(m+n+m-n).(m+n-m+n)`
`=2m.2n`
`=4.m.n`
`c)`
`C=a^4.b^2-x^8`
`=(a^2.b)^2-(x^4)^2`
`=(a^2.b-x^4).(a^2.b+x^4)`
`d)`
`D=(2a+b)^2-(2b+a)^2`
`=(2a+b-2b-a).(2a+b+2b+a)`
`=(a-b).(3a+3b)`
`=3.(a+b).(a-b)`
`e)`
`E=(a+b)^2-(b+c)^2`
`=(a+b+b+c).(a+b-b-c)`
`=(a+2b+c).(a-c)`
Bài 3:
`a)`
`A=x^3-y^3`
`=(x-y).(x^2+xy+y^2)`
`b)`
`B=x^3+27`
`=x^3+3^3`
`=(x+3).(x^2-3x+9)`
`c)`
`C=a^3-27`
`=a^3-3^3`
`=(a-3).(a^2+3a+9)`
`d)`
`D=125-b^3`
`=5^3-b^3`
`=(5-b).(25+5b+b^2)`
`e)`
`E=m^3-64`
`=m^3-4^3`
`=(m-4).(m^2+4m+16)`
Bài 4:
`a)`
`A=27a^3-8`
`=(3a)^3-2^3`
`=(3a-2).(9a^2+6a+4)`
`b)`
`B=(8a^3-27b^3)-2a(4a^2-9b^2)`
`=[(2a)^3-(3b)^3]-2a.[(2a)^2-(3b)^2]`
`=(2a-3b).(4a^2+6ab+9b^2)-2a.(2a+3b).(2a-3b)`
`=(2a-3b).(4a^2+6ab+9b^2-2a.(2a+3b))`
`=(2a-3b).(4a^2+6ab+9b^2-4a^2-6ab)`
`=(2a-3b).9.b^2`
`c)`
`C=(a^3-b^3)+(a-b)^2`
`=(a-b).(a^2+ab+b^2)+(a-b).(a-b)`
`=(a-b).(a^2+ab+b^2+a-b)`
`d)`
`D=(a^3+b^3)+(a+b)^2`
`=(a+b).(a^2-ab+b^2)+(a+b).(a+b)`
`=(a+b).(a^2-ab+b^2+a+b)`
Bài 5:
`a)`
`A=(a^2+1)^2-4a^2`
`=(a^2+1)^2-(2a)^2`
`=(a^2+1+2a).(a^2+1-2a)`
`=(a+1)^2.(a-1)^2`
`b)`
`B=(x^2+4)^2-16x^2`
`=(x^2+4)^2-(4x)^2`
`=(x^2-4x+4).(x^2+4x+4)`
`=(x-2)^2.(x+2)^2`
`c)`
`C=(a^2+2ab+b^2)-c^2`
`=(a+b)^2-c^2`
`=(a+b-c).(a+b+c)`
`d)`
`D=1-(x^2-2xy+y^2)`
`=1-(x-y)^2`
`=(1-x+y).(1+x-y)`
`e)`
`E=2x^2+2y^2-4xy`
`=2.(x^2-2xy+y^2)`
`=2.(x-y)^2`
Bài 6:
`a)`
`x^2-36=0`
`<=>x^2-6^2=0`
`<=>(x-6).(x+6)=0`
`<=>`\(\left[ \begin{array}{l}x-6=0\\x+6=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=6\\x=-6\end{array} \right.\)
Vậy `S={-6;6}`
`b)`
`4x^2+4x+1=0`
`<=>(2x)^2+2.2.x+1=0`
`<=>(2x+1)^2=0`
`<=>2x+1=0`
`<=>x=-1/2`
Vậy `S={-1/2}`
`c)`
`25-10x+x^2=0`
`5^2-2.5.x+x^2=0`
`<=>(5-x)^2=0`
`<=>5-x=0`
`<=>x=5`
Vậy `S={5}`
`d)`
`x^3+8=0`
`<=>x^3+2^3=0`
`<=>(x+2).(x^2-2x+4)=0`
`<=>`\(\left[ \begin{array}{l}x+2=0\\x^2-2x+4=0\end{array} \right.\)
`+)x^2-2x+4=(x-1)^2+3`
Vì `(x-1)^2≥0⇒(x-1)^2+3≥3>0`
`=>` Không có `x` thỏa mãn: `x^2-2x+4=0`
`+)x+2=0`
`<=>x=-2`
Vậy `S={-2}`
Bài 7:
`a)`
`x^3-3x^2+3x-1=0`
`<=>(x-1)^3=0`
`<=>x-1=0`
`<=>x=1`
Vậy `S={1}`
`b)`
`4x^3-36x=0`
`<=>4.x.(x^2-9)=0`
`<=>4.x.(x+3).(x-3)=0`
`<=>`\(\left[ \begin{array}{l}x=0\\x+3=0\\x-3=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=0\\x=-3\\x=3\end{array} \right.\)
Vậy `S={0;-3;3}`
`c)`
`x^6-1=0`
`<=>(x^2)^2-1^3=0`
`<=>(x^2-1).(x^4+x^2+1)=0`
Vì `x^4+x^2+1\ne0`
`<=>x^2-1=0`
`<=>(x-1).(x+1)=0`
`<=>`\(\left[ \begin{array}{l}x-1=0\\x+1=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\)
Vậy `S={-1;1}`
`d)`
`x^3-6x^2+12x-8=0`
`<=>x^3-3.x.x^2+3.2^2.x-2^3=0`
`<=>(x-2)^3=0`
`<=>x-2=0`
`<=>x=2`
Vậy `S={2}`
Bài 8:
`a)`
`9x^2-16.(x-1)^2=0`
`<=>(3x)^2-[4.(x-1)]^2=0`
`<=>[3x-4.(x-1)][3x+4.(x-1)]=0`
`<=>(3x-4x+4).(3x+4x-4)=0`
`<=>(4-x).(7x-4)=0`
`<=>`\(\left[ \begin{array}{l}4-x=0\\7x-4=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=4\\x=\dfrac{4}{7}\end{array} \right.\)
Vậy `S={4; 4/7}`
`b)`
`(5x-4)^2-49x^2=0`
`<=>(5x-4)^2-(7x)^2=0`
`<=>(5x-4-7x).(5x-4+7x)=0`
`<=>(-2x-4).(12x-4)=0`
`<=>`\(\left[ \begin{array}{l}-2x-4=0\\12x-4=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=-2\\x=\dfrac{1}{3}\end{array} \right.\)
Vậy `S={-2;1/3}`
`e)`
`(x^3+9)-(x+2).(x-4)`
Sai đề ; sửa:
`(x^3+8)-(x+2).(x-4)=0`
`<=>(x^3+2^3)-(x+2).(x-4)=0`
`<=>(x+2).(x^2-2x+4)-(x+2).(x-4)=0`
`<=>(x+2).(x^2-2x+4-x+4)=0`
`<=>(x+2).(x^2-3x+8)=0`
Vì `x^2-3x+8=x^2-2.(3)/(2).x+(9)/(4)+(23)/(4)=(x-3/2)^2+(23)/(4)≥23/4>0∀x`
`<=>x+2=0`
`<=>x=-2`
Vậy `S={-2}`
`b)` (trùng với `b)` bài `7`)
`4x^3-36x=0`
`<=>4.x.(x^2-9)=0`
`<=>4.x.(x+3).(x-3)=0`
`<=>`\(\left[ \begin{array}{l}x=0\\x+3=0\\x-3=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=0\\x=-3\\x=3\end{array} \right.\)
Vậy `S={0;-3;3}`
`d)` (trùng với `d)` bài `7`)
`x^3-6x^2+12x-8=0`
`<=>x^3-3.x.x^2+3.2^2.x-2^3=0`
`<=>(x-2)^3=0`
`<=>x-2=0`
`<=>x=2`
Vậy `S={2}`
