Question

P = (\(\sqrt{x}\) - \(\dfrac{x+2}{\sqrt{x}+1}\)) : (\(\dfrac{\sqrt{x}}{\sqrt{x}+1}\) - \(\dfrac{\sqrt{x}-4}{1-x}\))

a) Rút gọn P

b) Tìm x để P<0

Answer 1

đk x >= 0 ; x khác 1 

a, \(P=\left(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}-4}{x-1}\right)\)

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}:\dfrac{x-4}{x-1}=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)

b, Ta có P < 0 => \(\sqrt{x}-1< 0\Leftrightarrow x< 1\)

Kết hợp đk vậy 0 =< x < 1 

Answer 2

\(a,đkx>0;x\ne1\\ P=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)-x-2}{\sqrt{x}+1}\right)\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-4}{x-1}\right)\\ =\left(\dfrac{x+\sqrt{x}-x-2}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}-4}{x-1}\right)\\ =\dfrac{\sqrt{x}-2}{\sqrt{x}+1}:\dfrac{x-\sqrt{x}+\sqrt{x}-4}{x-1}\\ =\dfrac{\sqrt{x}-2}{\sqrt{x}+1}.\dfrac{x-1}{x-4}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)

\(b,P< 0\\ \dfrac{\sqrt{x}-1}{\sqrt{x}+2}< 0\\ \sqrt{x}-1< 0\\ x< 1\\ KHvsđk\\ 0< x< 1\)