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Ta có: \(\frac{x-2\sqrt{x}+8}{x-4}-\frac{2}{\sqrt{x}-2}\)
\(=\frac{x-2\sqrt{x}+8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x-2\sqrt{x}+8-2\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x-4\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}-2}{\sqrt{x}+2}\)
Ta có: \(\dfrac{x+\sqrt{x}}{\sqrt{x}}+\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+1+\sqrt{x}+2\)
\(=2\sqrt{x}+3\)
ĐKXĐ: \(x\ge-\dfrac{5}{2}\)
\(\sqrt{2x+5}+\sqrt{x+7}+x-8=0\\ \Leftrightarrow\left(\sqrt{2x+5}-3\right)+\left(\sqrt{x+7}-3\right)+x-2=0\\ \Leftrightarrow\dfrac{2x-4}{\sqrt{2x+5}+3}+\dfrac{x-2}{\sqrt{x+7}+3}+x-2=0\)
\(\Leftrightarrow\dfrac{2\left(x-2\right)}{\sqrt{2x+5}+3}+\dfrac{x-2}{\sqrt{x+7}+3}+x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{2}{\sqrt{2x+5}+3}+\dfrac{1}{\sqrt{x+7}+3}+1\right)=0\)
Vì \(\dfrac{2}{\sqrt{2x+5}+3}>0;\dfrac{1}{\sqrt{x+7}+3}>0;1>0\Rightarrow\dfrac{2}{\sqrt{2x+5}+3}+\dfrac{1}{\sqrt{x+7}+3}+1>0\)
\(\Rightarrow x-2=0\\ \Rightarrow x=2\left(tm\right)\)
Vậy \(x=2\)