Cho biết \(\left(\dfrac{4\sqrt{x}}{2+\sqrt{x}}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)=-1\). Hỏi x = ?
\(x=\dfrac{9}{16}\). \(x=\dfrac{3}{4}\). \(x=\dfrac{3}{4};x=-\dfrac{3}{4}\). \(x=-\dfrac{9}{16}\). Hướng dẫn giải:\(\left(\dfrac{4\sqrt{x}}{2+\sqrt{x}}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
\(=\left(\dfrac{4\sqrt{x}\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\dfrac{4x+8\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\dfrac{-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4\sqrt{x}}{2-\sqrt{x}}.\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{-\sqrt{x}+3}\)
\(=\dfrac{-4x}{-\sqrt{x}+3}\)
\(\dfrac{-4x}{-\sqrt{x}+3}=-1\)
\(\Leftrightarrow4x=-\sqrt{x}+3\)
\(\Leftrightarrow4x+\sqrt{x}-3=0\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(4\sqrt{x}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=0\\4\sqrt{x}-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-1\left(l\right)\\\sqrt{x}=\dfrac{3}{4}\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{9}{16}\)