Question

Bài 8. Thực hiện phép tính

a)( \(\dfrac{1}{x^2+x}\) + \(\dfrac{2-x}{x+1}\) ) : ( \(\dfrac{1}{x}\) + x - 2)

Answer 1

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

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

Answer 2

`ĐK x \ne -1,x \ne 0`

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

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

Answer 3

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

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

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

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

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

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