Đáp án:

Giải thích các bước giải:

 a)$(\frac{2}{x-2}-\frac{2}{x+2}).\frac{x^{2}+4x+4}{8}$

= $\frac{2(x+2)-2(x-2)}{(x-2)(x+2)}.\frac{(x+2)^{2}}{8}$

= $\frac{2x+4-(2x-4)}{(x-2)(x+2)}.\frac{(x+2)^{2}}{8}$ 

= $\frac{8}{(x-2)(x+2)}.\frac{(x+2)^{2}}{8}$ 

= $\frac{x+2}{x-2}$

b) $\frac{x+2}{4x + 24}.\frac{x^{2}-36}{x^{2}+x-2}$

= $\frac{x+2}{4(x+6)}.\frac{(x-6)(x+6)}{(x+2)(x-1)}$

= $\frac{x-6}{4(x-1)}$ 

2)

a)$(x^{2}-5):\frac{2x+10}{3x-7}$

= $(x-5)(x+5).\frac{3x-7}{2(x+5)}$

= $\frac{(x-5)(3x-7)}{2}$ 

b)$(\frac{3x}{1-3x}+\frac{2x}{3x+1}):\frac{6x^{2}+10}{1-6x+9x^{2}}$ 

= $\frac{3x(3x+1)+2x(1-3x)}{(3x+1)(1-3x)}.\frac{(3x-1)^{2}}{2(3x^{2}+5)}$

= $\frac{3x^{2}+2x}{(3x+1)(1-3x)}.\frac{(3x-1)^{2}}{2(3x^{2}+5)}$

= $\frac{x(3x+2)}{-(3x+1)(3x-1)}.\frac{(3x-1)^{2}}{2(3x^{2}+5)}$

= $\frac{x(3x+2)(3x-1)}{-2(3x+1)(3x^{2}+5)}$

Đáp án:

Giải thích các bước giải:

 a)(2x−2−2x+2).x2+4x+48(2x−2−2x+2).x2+4x+48

= 2(x+2)−2(x−2)(x−2)(x+2).(x+2)282(x+2)−2(x−2)(x−2)(x+2).(x+2)28

= 2x+4−(2x−4)(x−2)(x+2).(x+2)282x+4−(2x−4)(x−2)(x+2).(x+2)28 

= 8(x−2)(x+2).(x+2)288(x−2)(x+2).(x+2)28 

= x+2x−2x+2x−2

b) x+24x+24.x2−36x2+x−2x+24x+24.x2−36x2+x−2

= x+24(x+6).(x−6)(x+6)(x+2)(x−1)x+24(x+6).(x−6)(x+6)(x+2)(x−1)

= x−64(x−1)x−64(x−1) 

2)

a)(x2−5):2x+103x−7(x2−5):2x+103x−7

= (x−5)(x+5).3x−72(x+5)(x−5)(x+5).3x−72(x+5)

= (x−5)(3x−7)2(x−5)(3x−7)2 

b)(3x1−3x+2x3x+1):6x2+101−6x+9x2(3x1−3x+2x3x+1):6x2+101−6x+9x2 

= 3x(3x+1)+2x(1−3x)(3x+1)(1−3x).(3x−1)22(3x2+5)3x(3x+1)+2x(1−3x)(3x+1)(1−3x).(3x−1)22(3x2+5)

= 3x2+2x(3x+1)(1−3x).(3x−1)22(3x2+5)3x2+2x(3x+1)(1−3x).(3x−1)22(3x2+5)

= x(3x+2)−(3x+1)(3x−1).(3x−1)22(3x2+5)x(3x+2)−(3x+1)(3x−1).(3x−1)22(3x2+5)

= x(3x+2)(3x−1)−2(3x+1)(3x2+5)