a) ĐKXĐ: $x-1\geq0$ $x<=>$ $x\geq1$
$P=(\frac{1}{x-1}+\frac{1}{\sqrt{x}+1}):\frac{1}{x+\sqrt{x}}$
$=(\frac{1}{(\sqrt{x}-1)(\sqrt{x}+1)}+\frac{\sqrt{x}-1}{(\sqrt{x}-1)(\sqrt{x}+1)}).{\sqrt{x}(\sqrt{x}+1)}$
$=\frac{\sqrt{x}}{(\sqrt{x}-1)(\sqrt{x}+1)}.{\sqrt{x}(\sqrt{x}+1)}$
$=\frac{x}{\sqrt{x}-1}$
b) Thay $x=\frac{1}{16}$ vào P, ta có:
$P=\frac{\frac{1}{16}}{\sqrt{\frac{1}{16}}-1}=\frac{\frac{1}{16}}{\frac{1}{4}-1}=\frac{1}{16}:\frac{-3}{4}=\frac{1}{16}.\frac{-4}{3}=\frac{-1}{12}$
c) Để $P=2\sqrt{x}$ thì:
$\frac{x}{\sqrt{x}-1}=2\sqrt{x}$
$<=>\frac{x}{\sqrt{x}-1}=\frac{2x-2\sqrt{x}}{\sqrt{x}-1}$
$<=>x=2x-2\sqrt{x}$
$<=>x-2x+2\sqrt{x}=0$
$<=>2\sqrt{x}-x=0$
$<=>\sqrt{x}(2-\sqrt{x})=0$
$<=>\left[ \begin{array}{l}\sqrt{x}=0\\2-\sqrt{x}=0\end{array} \right.$
$<=>\left[ \begin{array}{l}x=0\\\sqrt{x}=2\end{array} \right.$
$<=>\left[ \begin{array}{l}x=0(loại)\\x=4(nhận)\end{array} \right.$
Vậy $x=4$
#NOCOPY
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