Question

Cho : A =( \(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{1}{x-\sqrt{x}}\)) +\(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\)(x < 0 ; x ≠ 1 )

a, Rút gọn

b, 1; tim x 2; c/m A < 0

Answer 1

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

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

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

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

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

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

b: Để A<0 thì \(\sqrt{x}-1< 0\)

=>0<x<1