Question

Cho A = \(\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)

a) Rút gọn A.

b) Tìm x để A > 0

Answer 1

\(A=\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{\left(1-x\right)\left(1+x\right)}\right):\dfrac{1-2x}{x^2-1}\)

\(A=\dfrac{x+1+2-2x-5+x}{\left(1-x\right)\left(1+x\right)}:\dfrac{1-2x}{x^2-1}\)

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

\(A=\dfrac{-2x+2}{1-2x}\)

\(A=0\)

⇔\(\dfrac{-2x+2}{1-2x}>0\)

⇔\(-2x+1>0\)

⇔\(-2x>-1\)

⇔\(x< \dfrac{1}{2}\)

Vậy x<\(\dfrac{1}{2}\) thì A>0