ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)

Câu a :

\(P=\left(\dfrac{4\sqrt{x}}{2+\sqrt{x}}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)

\(=\left(\dfrac{4\sqrt{x}\left(2-\sqrt{x}\right)+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)

\(=\dfrac{8\sqrt{x}+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{4\sqrt{x}}{2-\sqrt{x}}\times\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)

\(=\dfrac{-4x}{3-\sqrt{x}}=\dfrac{4x}{\sqrt{x}-3}\)

Câu b :

\(P=-1\)

\(\Leftrightarrow\) \(\dfrac{4x}{\sqrt{x}-3}=-1\)

\(\Leftrightarrow4x=-\sqrt{x}+3\)

\(\Leftrightarrow4x+\sqrt{x}-3=0\)

\(\Leftrightarrow4x+4\sqrt{x}-3\sqrt{x}-3=0\)

\(\Leftrightarrow4\sqrt{x}\left(\sqrt{x}+1\right)-3\left(\sqrt{x}+1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(4\sqrt{x}-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=0\\4\sqrt{x}-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\x=\dfrac{9}{16}\end{matrix}\right.\)

Vậy \(x=\dfrac{9}{16}\)

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