`a)` `P=(1/{\sqrt{x}+2}+7/{x-4}):({\sqrt{x}-1}/{\sqrt{x}-2}-1)`
`\qquad (x\ge 0;x\ne 4)`
`={\sqrt{x}-2+7}/{(\sqrt{x}+2)(\sqrt{x}-2)}:{\sqrt{x}-1-(\sqrt{x}-2)}/{\sqrt{x}-2}`
`={\sqrt{x}+5}/{(\sqrt{x}+2)(\sqrt{x}-2)}: 1/{\sqrt{x}-2}`
`={\sqrt{x}+5}/{(\sqrt{x}+2)(\sqrt{x}-2)} . (\sqrt{x}-2)`
`={\sqrt{x}+5}/{\sqrt{x}+2}`
Vậy `P={\sqrt{x}+5}/{\sqrt{x}+2}` với `x\ge 0;x\ne 4`
$\\$
`b)`
`1)`Với `x=9` (thỏa mãn đk)
`=>P={\sqrt{9}+5}/{\sqrt{9}+2}={3+5}/{3+2}=8/5`
Vậy `P=8/5` với `x=9`
$\\$
+) Với `x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}`
`=\sqrt{5^2+2.5.\sqrt{2}+2}-\sqrt{4^2+2.4.\sqrt{2}+2}`
`=\sqrt{(5+\sqrt{2})^2}-\sqrt{(4+\sqrt{2})^2}`
`=5+\sqrt{2}-(4+\sqrt{2})=1` (thỏa mãn)
`=>P={\sqrt{1}+5}/{\sqrt{1}+2}=6/3=2`
Vậy `P=2` với `x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}``
$\\$
`3)` `x=\sqrt{2/{2-\sqrt{3}}}-\sqrt{2/{2+\sqrt{3}}}`
`=\sqrt{{2(2+\sqrt{3})}/{(2-\sqrt{3})(2+\sqrt{3})}}-\sqrt{{2(2-\sqrt{3})}/{(2+\sqrt{3})(2-\sqrt{3})}}`
`=\sqrt{{4+2\sqrt{3}}/{2^2-3}}-\sqrt{{4-2\sqrt{3}}/{2^2-3}}`
`=\sqrt{3+2\sqrt{3}.1+1^2}-\sqrt{3-2\sqrt{3}.1+1^2}`
`=\sqrt{(\sqrt{3}+1)^2}-\sqrt{(\sqrt{3}-1)^2}`
`=|\sqrt{3}+1|-|\sqrt{3}-1|=\sqrt{3}+1-(\sqrt{3}-1)=2` (thỏa mãn)
`=>P={\sqrt{2}+5}/{\sqrt{2}+2}`
`={(\sqrt{2}+5)(\sqrt{2}-2)}/{(\sqrt{2}+2)(\sqrt{2}-2)}`
`={2-2\sqrt{2}+5\sqrt{2}-10}/{2-4}={3\sqrt{2}-8}/{-2}`
`={8-3\sqrt{2}}/2`
$\\$
_________
`3)` `x=\sqrt{2}/{2-\sqrt{3}}-\sqrt{2}/{2+\sqrt{3}}`
`=\sqrt{2}. (1/{2-\sqrt{3}}-1/{2+\sqrt{3}})`
`=\sqrt{2}.{2+\sqrt{3}-(2-\sqrt{3})}/{(2-\sqrt{3}).(2+\sqrt{3})}`
`=\sqrt{2}. {2\sqrt{3}}/{2^2-3}=2\sqrt{6}` (thỏa mãn)
`=>P={\sqrt{2\sqrt{6}}+5}/{\sqrt{2\sqrt{6}}+2}`
