\(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{x-1}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)\(\left(đk:x\ne1;x\ge0\right)\)
\(=\dfrac{\sqrt{x}+1+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)
\(=\dfrac{1}{\sqrt{x}+1}\)
Để P nguyên
=> \(\dfrac{1}{\sqrt{x}+1}\in Z\)
\(\Rightarrow\left\{{}\begin{matrix}\text{x là số chính phương}\\\sqrt{x}+1\inƯ\left(1\right)=\left\{1;-1\right\}\end{matrix}\right.\)
\(\Rightarrow x=0\)
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