1, ĐKXĐ: \(x>0;x\ne1\)
\(P=\frac{3x+3\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\frac{\sqrt{x}-2}{\sqrt{x}}\cdot\frac{1-1+\sqrt{x}}{1-\sqrt{x}}\\ =\frac{3x+3\sqrt{x}-3-x-1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{\sqrt{x}-2}{\sqrt{x}-1}\\ =\frac{2x+3\sqrt{x}-4-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\frac{2x+3\sqrt{x}-4-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\\ =\frac{x+3\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
2, Để \(P=\sqrt{x}\) thì:
\(\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\sqrt{x}\\ \Leftrightarrow\sqrt{x}\left(\sqrt{x}+3\right)=\sqrt{x}\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)\\ \Leftrightarrow\sqrt{x}+3=x+\sqrt{x}-2\\ \Leftrightarrow x-5=0\Leftrightarrow x=5\left(t/m\right)\)
Vậy với \(x=5\) thì \(P=\sqrt{x}\).
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