a) ĐKXĐ: x\(\ne1,x\ge0\)
\(P=\left(\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}+\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+2}\right):\dfrac{1}{x-1}=\left(\dfrac{3x+3\sqrt{x}-3+\sqrt{x}+2+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=\left(\dfrac{\left(3x+5\sqrt{x}-2\right)\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right)=\dfrac{\left(3\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+2}=3x+3\sqrt{x}-\sqrt{x}-1=3x+2\sqrt{x}-1\)b)Ta có \(\dfrac{1}{P}=\dfrac{1}{3x+2\sqrt{x}-1}=\dfrac{1}{\left(3\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
Để \(\dfrac{1}{P}\in N\) thì \(\left(3\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\inƯ\left(1\right)\)\(\in\left(\pm1\right)\)
Ta có \(\sqrt{x}+1>0\Rightarrow3\sqrt{x}-1>0\)
Mà 1 là số nguyên tố \(\Rightarrow\left\{{}\begin{matrix}3\sqrt{x}-1=1\\\sqrt{x}+1=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{4}{9}\\x=0\end{matrix}\right.\)
(loại)
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