Đáp án:
`a)` `M=2/3` khi `x=9`
`b)` `P={\sqrt{x}+6}/{\sqrt{x}-1}`
`c)` `Q>3`
Giải thích các bước giải:
`a)` `M={\sqrt{x}-1}/\sqrt{x}` `\quad (x>0)`
Với `x=9` (thỏa mãn)
`=>M={\sqrt{9}-1}/\sqrt{9}=2/3`
Vậy `M=2/3` khi `x=9`
$\\$
`b)` `P={\sqrt{x}-2}/{\sqrt{x}+1}+{2+8\sqrt{x}}/{x-1}-2/{1-\sqrt{x}}`
`\qquad (x>0;x\ne 1;x\ne 5)`
`={\sqrt{x}-2}/{\sqrt{x}+1}+{2+8\sqrt{x}}/{(\sqrt{x}-1)(\sqrt{x}+1)}+2/{\sqrt{x}-1}`
`={(\sqrt{x}-2)(\sqrt{x}-1)+2+8\sqrt{x}+2(\sqrt{x}+1)}/{(\sqrt{x}-1)(\sqrt{x}+1)}`
`={x-\sqrt{x}-2\sqrt{x}+2+2+8\sqrt{x}+2\sqrt{x}+2}/{(\sqrt{x}-1)(\sqrt{x}+1)}`
`={x+7\sqrt{x}+6}/{(\sqrt{x}-1)(\sqrt{x}+1)}`
`={x+\sqrt{x}+6\sqrt{x}+6}/{(\sqrt{x}-1)(\sqrt{x}+1)}`
`={\sqrt{x}(\sqrt{x}+1)+6(\sqrt{x}+1)}/{(\sqrt{x}-1)(\sqrt{x}+1)}`
`={(\sqrt{x}+6)(\sqrt{x}+1)}/{(\sqrt{x}-1)(\sqrt{x}+1)}={\sqrt{x}+6}/{\sqrt{x}-1}`
Vậy `P={\sqrt{x}+6}/{\sqrt{x}-1}` (đpcm)
$\\$
`c)` `Q=M. P+{x-5}/\sqrt{x}`
`={\sqrt{x}-1}/\sqrt{x} . {\sqrt{x}+6}/{\sqrt{x}-1} +{x-5}/\sqrt{x}`
`={\sqrt{x}+6}/\sqrt{x}+{x-5}/\sqrt{x}`
`={\sqrt{x}+6+x-5}/\sqrt{x}={x+\sqrt{x}+1}/\sqrt{x}`
`Q-3={x+\sqrt{x}+1}/\sqrt{x}-3`
`={x+\sqrt{x}+1-3\sqrt{x}}/\sqrt{x}`
`={x-2\sqrt{x}+1}/\sqrt{x}={(\sqrt{x}-1)^2}/\sqrt{x}`
Với mọi `x>0;x\ne 1;x\ne 5` ta có:
`\qquad (\sqrt{x}-1)^2>0`
`\qquad \sqrt{x}>0`
`=>{(\sqrt{x}-1)^2}/\sqrt{x}>0`
`=>Q-3>0=>Q>3`
Vậy `Q>3`
