Đáp án: `a)P=\frac{-x-2\sqrt{x}+2}{\sqrt{x}}(x≥1;x\ne2;x\ne3)`
`b)P=1+\sqrt{2}` với $x=3-2\sqrt{2}$
Giải thích các bước giải:
$a)ĐKXĐ:x≥1;x\neq2;x\neq3$
`P=(\frac{1}{\sqrt{x}-\sqrt{x-1}}-\frac{x-3}{\sqrt{x-1}-\sqrt{2}})(\frac{2}{\sqrt{2}-\sqrt{x}}-\frac{\sqrt{x}+\sqrt{2}}{\sqrt{2x}-x})`
`=(\frac{\sqrt{x-1}+\sqrt{x}}{x-(x-1)}-\frac{x-1-2}{\sqrt{x-1}-\sqrt{2}}).\frac{2\sqrt{x}-(\sqrt{x}+\sqrt{2})(\sqrt{2}-\sqrt{x})}{\sqrt{x}(\sqrt{2}-\sqrt{x})}`
`=[\sqrt{x-1}+\sqrt{x}-\frac{(\sqrt{x-1}-\sqrt{2})(\sqrt{x-1}+\sqrt{2})}{\sqrt{x-1}-\sqrt{2}}].\frac{2\sqrt{x}-(2-x)}{\sqrt{x}(\sqrt{2}-\sqrt{x})}`
`=[\sqrt{x-1}+\sqrt{x}-(\sqrt{x-1}+\sqrt{2})].\frac{2\sqrt{x}-2+x}{\sqrt{x}(\sqrt{2}-\sqrt{x})}`
`=(\sqrt{x-1}+\sqrt{x}-\sqrt{x-1}-\sqrt{2}).\frac{-x-2\sqrt{x}+2}{\sqrt{x}(\sqrt{x}-\sqrt{2})}`
`=(\sqrt{x}-\sqrt{2}).\frac{-x-2\sqrt{x}+2}{\sqrt{x}(\sqrt{x}-\sqrt{2})}=\frac{-x-2\sqrt{x}+2}{\sqrt{x}}`
$b)$ Ta có: $x=3-2\sqrt{2}$ (thỏa mãn)
$⇒\sqrt{x}=\sqrt{3-2\sqrt{2}}=\sqrt{(\sqrt{2}-1)^2}=|\sqrt{2}-1|=\sqrt{2}-1$
Thay $x;\sqrt{x}$ vào $P$ ta được:
`P=\frac{-(3-2\sqrt{2})-2(\sqrt{2}-1)+2}{\sqrt{2}-1}=\frac{-3+2\sqrt{2}-2\sqrt{2}+2+2}{\sqrt{2}-1}`
`=\frac{1}{\sqrt{2}-1}=\frac{\sqrt{2}+1}{2-1}=1+\sqrt{2}`
