ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne\pm1\end{matrix}\right.\)
Ta có :
\(P=\left(\frac{x+\sqrt{x}+1}{x+\sqrt{x}-2}+\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+2}\right):\frac{1}{x-1}\)
\(=\left(\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+2}\right):\frac{1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\left(\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right).\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\)
\(=\frac{x+\sqrt{x}+1+\sqrt{x}+2+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}.\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(=\frac{x+3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}.\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}.\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(=\left(\sqrt{x}+1\right)^2\)
Vậy..
