a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
Ta có: \(A=\left(\sqrt{x}-\dfrac{x+2}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-4}{x-1}\right)\)
\(=\dfrac{x-\sqrt{x}-x-2}{\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x-\sqrt{x}+\sqrt{x}-4}\)
\(=\dfrac{-\left(\sqrt{x}+2\right)\cdot\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{-\sqrt{x}-1}{\sqrt{x}-2}\)
a) `ĐK: x >= 0 ;x \ne 1`
\(A=\left(\sqrt{x}-\dfrac{x+2}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-4}{x-1}\right)\\ =\dfrac{x-\sqrt{x}-x-2}{\sqrt{x}-1}:\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{-\sqrt{x}-2}{\sqrt{x}-1}.\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{x-4}\\ =\dfrac{-\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\\ =\dfrac{-\sqrt{x}-1}{\sqrt{x}-2}\)
b)
Khi `x=9 => A=(-\sqrt9-1)/(\sqrt9-2)=-4`
Khi `x=1/4 => A=(-\sqrt(1/4)-1)/(\sqrt(1/4)-2)=1`
c) `A<1/2 <=> (-\sqrtx-1)/(\sqrtx-2) <1/2`
`<=>(-\sqrtx-1-2\sqrtx+4)/(\sqrtx-2)<0`
`<=>(-3\sqrtx+3)/(\sqrtx-2)<0`
`<=>(\sqrtx-1)/(\sqrtx-2)>0`
TH1: `{(\sqrtx-1>0),(\sqrtx-2>0):} <=> x>4`
TH2: `{(\sqrtx-1<0),(\sqrtx-2<0):} <=>x<1`
Vậy `0<= x <1 \vee x>4` thỏa mãn.
=x−√x−x−2(√x−1)⋅(√x+1)(√x−1)x−√x+√x−4=x−x−x−2(x−1)⋅(x+1)(x−1)x−x+x−4
=−√x−1√x−2
