ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
\(\dfrac{x-1}{4\sqrt{x}}\cdot\dfrac{2x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{13}{8}\)
=>\(\dfrac{\left(x-1\right)}{4\sqrt{x}}\cdot\dfrac{2x+2\sqrt{x}+1}{x-1}=\dfrac{13}{8}\)
=>\(\dfrac{2x+2\sqrt{x}+1}{4\sqrt{x}}=\dfrac{13}{8}\)
=>\(8\left(2x+2\sqrt{x}+1\right)=13\cdot4\sqrt{x}=52\sqrt{x}\)
=>\(16x+16\sqrt{x}+16-52\sqrt{x}=0\)
=>\(16x-36\sqrt{x}+16=0\)
=>\(4x-9\sqrt{x}+4=0\)
=>\(x-\dfrac{9}{4}\sqrt{x}+1=0\)
=>\(x-2\cdot\sqrt{x}\cdot\dfrac{9}{8}+\dfrac{81}{64}-\dfrac{17}{64}=0\)
=>\(\left(\sqrt{x}-\dfrac{9}{8}\right)^2-\dfrac{17}{64}=0\)
=>\(\left(\sqrt{x}-\dfrac{9}{8}-\dfrac{\sqrt{17}}{8}\right)\left(\sqrt{x}-\dfrac{9}{8}+\dfrac{\sqrt{17}}{8}\right)=0\)
=>\(\left[{}\begin{matrix}\sqrt{x}-\dfrac{9}{8}-\dfrac{\sqrt{17}}{8}=0\\\sqrt{x}-\dfrac{9}{8}+\dfrac{\sqrt{17}}{8}=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}\sqrt{x}=\dfrac{9+\sqrt{17}}{8}\\\sqrt{x}=\dfrac{9-\sqrt{17}}{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{49+9\sqrt{17}}{32}\left(nhận\right)\\x=\dfrac{49-9\sqrt{17}}{32}\left(nhận\right)\end{matrix}\right.\)
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