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a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/ ĐKXĐ: \(\hept{\begin{cases}x>0\\x\ne4\end{cases}}\)
\(A=\left[\frac{x}{\sqrt{x}\left(x-4\right)}-\frac{6}{3\left(\sqrt{x}-2\right)}+\frac{1}{\sqrt{x}-2}\right]:\left(\frac{x-4+10-x}{\sqrt{x}+2}\right)\)
\(=\left[\frac{\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}-2}+\frac{1}{\sqrt{x}-2}\right]:\left(\frac{6}{\sqrt{x}+2}\right)\)
\(=\frac{\sqrt{x}-2\left(\sqrt{x}+2\right)+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\frac{\left(\sqrt{x}+2\right)}{6}\)
\(=\frac{-2}{\sqrt{x}-2}.\frac{1}{6}=-\frac{1}{3\left(\sqrt{x}-2\right)}\)
2/ Để \(A>2\Rightarrow\frac{-1}{3\left(\sqrt{x}-2\right)}>2\)\(\Rightarrow6\sqrt{x}-12+1>0\Rightarrow6\sqrt{x}-11>0\Rightarrow\sqrt{x}>\frac{11}{6}\)
\(\Rightarrow x>\frac{121}{36}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(A=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)+2\sqrt{x}\left(\sqrt{x}+2\right)-3x-4}{x-4}\)
\(=\dfrac{x-2\sqrt{x}+2x+4\sqrt{x}-3x-4}{x-4}\)
\(=\dfrac{2\sqrt{x}-4}{x-4}=\dfrac{2}{\sqrt{x}+2}\)
b: A=1/2
=>\(\sqrt{x}+2=4\)
=>\(\sqrt{x}=2\)
=>x=4(loại)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(A=\dfrac{1}{\sqrt{x}+1}:\left(\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\dfrac{1}{\sqrt{x}+1}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
b: Để A<0 thì \(\sqrt{x}-2< 0\)
hay 0<x<4