Question

A= \(\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1+\dfrac{\sqrt{x}}{x+1}\right)\)
a) Rút gọn A
b) Tìm x để A ≤ 0

Answer 1

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

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

b: Để A<=0 thì -căn x+1<=0

=>-căn x<=-1

=>x>1