Question

Cho : \(\left(\dfrac{1}{x-1}-\dfrac{x}{1-x^3}.\dfrac{x^2+x+1}{x+1}\right):\dfrac{2x+1}{x^2+2x+1}\)

a. Rút gọn.

b, Tính GTBT khi \(x=\dfrac{1}{2}\)

Answer 1

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

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

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

b: Khi x=1/2thì \(P=\left(\dfrac{1}{2}+1\right):\left(\dfrac{1}{2}-1\right)=\dfrac{3}{2}:\dfrac{-1}{2}=-3\)