Question

thực hiện phép tính :
a) \(\left(\dfrac{1}{x^2+1}-\dfrac{2-x}{x+1}\right)\):\(\left(\dfrac{1}{x}+x-2\right)\)

Answer 1

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

\(=\dfrac{x+1+x^3+x-2x^2-2}{\left(x+1\right)\left(x^2+1\right)}\cdot\dfrac{x}{\left(x-1\right)^2}\)

\(=\dfrac{x^3-2x^2+2x-1}{\left(x+1\right)\left(x^2+1\right)}\cdot\dfrac{x}{\left(x-1\right)^2}\)

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

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

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