Question

Answer 1

Với `x \ne -1,x \ne 0` có:

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

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

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

`=1/[x+1]`

Answer 2

(1/x^2+x-(2-x)/x+1):(1/x+x-2)

\(=\left(\dfrac{1}{x\left(x+1\right)}+\dfrac{x-2}{x+1}\right):\dfrac{1+x^2-2x}{x}\)

\(=\dfrac{1+x^2-2x}{x\left(x+1\right)}\cdot\dfrac{x}{x^2-2x+1}=\dfrac{1}{x+1}\)