Lời giải:
ĐK:$x\geq 0; x\neq 1$
a) \(P=\frac{2+\sqrt{x}}{\sqrt{x}-1}.\frac{\sqrt{x}}{x+\sqrt{x}+2}=\frac{x+2\sqrt{x}}{(\sqrt{x}-1)(x+\sqrt{x}+2)}\)
b) \(x=3+2\sqrt{2}=2+1+2\sqrt{2.1}=(\sqrt{2}+1)^2\Rightarrow \sqrt{x}=\sqrt{2}+1\)
\(P=\frac{3+2\sqrt{2}+2\sqrt{2}+2}{(\sqrt{2}+1-1)(3+2\sqrt{2}+\sqrt{2}+1+2)}=\frac{5+4\sqrt{2}}{\sqrt{2}(6+3\sqrt{2})}=\frac{3+\sqrt{2}}{6}\)
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