Đáp án+Giải thích các bước giải:

`a, \frac{x^2-2x+1}{x^4}.\frac{x^2+2x}{x-1}`

`=\frac{(x-1)^2}{(x-2)(x+2)}.\frac{x(x+2)}{x-1}`

`=\frac{x(x-1)}{x-2}`

`=\frac{x^2-x}{x-2}`

`b, \frac{x^2-1}{x+5}.\frac{2x+10}{x^2-x}`

`=\frac{(x-1)(x+1)}{x+5}.\frac{2(x+5)}{x(x-1)}`

`=\frac{2(x+1)}{x}`

`=\frac{2x+2}{x}`

`c, \frac{x}{x+1}(\frac{x^3+1}{x^2-x+1}+\frac{x+1}{x})`

`=\frac{x}{x+1}(\frac{(x+1)(x^2-x+1)}{x^2-x+1}+\frac{x+1}{x})`

`=\frac{x}{x+1}(x+1+\frac{x+1}{x})`

`=\frac{x}{x+1}.\frac{x(x+1)+(x+1)}{x}`

`=\frac{x}{x+1}.\frac{x^2+x+x+1}{x}`

`=\frac{x}{x+1}.\frac{x^2+2x+1}{x}`

`=\frac{x}{x+1}.\frac{(x+1)^2}{x}`

`=x+1`

NẾU CÓ SAI ĐỀ THÌ BẢO MIK NHA TẠI CHẮC BẠN VIẾT VỘI QUÁ NÊN VIẾT ĐỀ KO RÕ MIK NHÌN LỘN

Đáp án:

`a)`

ĐK : `x \ne +-2; x \ne 1`

`(x^(2)-2x+1)/(x^(2)-4).(x^(2)+2x)/(x-1)`

`=(x-1)^2/[(x-2)(x+2)].[x(x+2)]/(x-1)`

`=[(x-1)^(2)x.(x+2)]/[(x-2)(x+2)(x-1)]`

`=[x(x-1)]/(x-2)`

`=(x^(2)-x)/(x-2)`

`b)`

ĐK : `x \ne 0; x \ne 1; x \ne -5`

`(x^(2)-1)/(x+5).(2x+10)/(x^(2)-x)`

`=[(x-1)(x+1)]/(x+5).[2(x+5)]/[x(x-1)]`

`=[(x-1)(x+1).2.(x+5)]/[(x+5).x(x-1)]`

`=[2(x+1)]/x`

`=(2x+2)/x`

`c)`

ĐK : `x \ne -1; x \ne 0`.

`x/(x+1).((x^(3)+1)/(x^(2)-x+1)+(x+1)/x)`

`=x/(x+1).([(x+1)(x^(2)-x+1)]/(x^(2)-x+1)+(x+1)/x)`

`=x/(x+1).(x+1+(x+1)/x)`

`=x/(x+1).((x^(2)+x+x+1)/x)`

`=x/(x+1).((x^(2)+2x+1)/x)`

`=x/(x+1).(x+1)^2/x`

`=x+1`