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ĐKXĐ: $x\geq 0$.
Ta có:
$2(x-\sqrt{x})^2-4(x-\sqrt{x})+4=\frac{3x+3\sqrt{x}-1}{(x+1)\sqrt{\sqrt{x}+1}}$
$\Leftrightarrow 2(x-\sqrt{x}-1)^2+\frac{2(x+1)\sqrt{\sqrt{x}+1}-3x-3\sqrt{x}+1}{(x+1)\sqrt{\sqrt{x}+1}}=0$
$\Leftrightarrow 2(x-\sqrt{x}-1)^2+\frac{(\sqrt{x}-\sqrt{\sqrt{x}+1})^2(1+2\sqrt{\sqrt{x}+1})}{2(x+1)\sqrt{\sqrt{x}+1}}=0$
$\Leftrightarrow \begin{cases} x-\sqrt{x}-1=0 \\ \sqrt{x}-\sqrt{\sqrt{x}+1}=0 \end{cases}$
$\Leftrightarrow x-\sqrt{x}-1=0$
$\Leftrightarrow x=\frac{3+\sqrt{5}}{2}$
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