Đáp án:

`a)` `P={\sqrt{a}-4}/{\sqrt{a}-2}` với `a\ge 0;a\ne 4`

`b)` `a>4`

Giải thích các bước giải:

`a)` `P={\sqrt{a}+2}/{\sqrt{a}+3}-5/{a+\sqrt{a}-6}+1/{2-\sqrt{a}}`

`\qquad (a\ge 0;a\ne 4)`

 `={sqrt{a}+2}/{\sqrt{a}+3}-5/{(\sqrt{a}-2)(\sqrt{a}+3)}-1/{\sqrt{a}-2}`

`={(\sqrt{a}+2)(\sqrt{a}-2)-5-1.(\sqrt{a}+3)}/{(\sqrt{a}-2)(\sqrt{a}+3)}`

`={a-4-5-\sqrt{a}-3}/{(\sqrt{a}-2)(\sqrt{a}+3)}`

`={a-\sqrt{a}-12}/{(\sqrt{a}-2)(\sqrt{a}+3)}``={a-4\sqrt{a}+3\sqrt{a}-12}/{(\sqrt{a}-2)(\sqrt{a}+3)}`

`={\sqrt{a}(\sqrt{a}-4)+3(\sqrt{a}-4)}/{(\sqrt{a}-2)(\sqrt{a}+3)}`

`={(\sqrt{a}-4)(\sqrt{a}+3)}/{(\sqrt{a}-2)(\sqrt{a}+3)}`

`={\sqrt{a}-4}/{\sqrt{a}-2}`

Vậy `P={\sqrt{a}-4}/{\sqrt{a}-2}` với `a\ge 0;a\ne 4`

$\\$

`b)` Để `P<1`

`<=>{\sqrt{a}-4}/{\sqrt{a}-2}<1`  (`a\ge 0;a\ne 4)`

`<=>{\sqrt{a}-4}/{\sqrt{a}-2}-1<0`  

`<=>{\sqrt{a}-4-(\sqrt{a}-2)}/{\sqrt{a}-2}<0`

`<=>{-2}/{\sqrt{a}-2}<0`

`<=>2/{\sqrt{a}-2}>0`

`<=>\sqrt{a}-2>0` 

`<=>\sqrt{a}>2`

`<=>(\sqrt{a})^2>2^2`

`<=>a>4` (thỏa mãn)

Vậy `a>4` thì `P<1`

`***`Lời giải`***`

a)

ĐKXĐ: `a≥0;ane4`

`P=\frac{\sqrt{a}+2}{\sqrt{a}+3}-\frac{5}{a+\sqrt{a}-6}+\frac{1}{2-\sqrt{a}} `

`=\frac{\sqrt{a}+2}{\sqrt{a}+3}-\frac{5}{a+3\sqrt{a}-2\sqrt{a}-6}-\frac{1}{\sqrt{a}-2} `

`=\frac{\sqrt{a}+2}{\sqrt{a}+3}-\frac{5}{\sqrt{a}(\sqrt{a}+3)-2(\sqrt{a}+3)}-\frac{1}{\sqrt{a}-2} `

`=\frac{\sqrt{a}+2}{\sqrt{a}+3}-\frac{5}{(\sqrt{a}-2)(\sqrt{a}+3)}-\frac{1}{\sqrt{a}-2} `

`=\frac{(\sqrt{a}+2)(\sqrt{a}-2)-5-(\sqrt{a}+3)}{(\sqrt{a}-2)(\sqrt{a}+3)} `

`=\frac{a-4-5-\sqrt{a}-3}{(\sqrt{a}-2)(\sqrt{a}+3)} `

`=\frac{a-\sqrt{a}-12}{(\sqrt{a}-2)(\sqrt{a}+3)} `

`=\frac{a+3\sqrt{a}-4\sqrt{a}-12}{(\sqrt{a}-2)(\sqrt{a}+3)} `

`=\frac{\sqrt{a}(\sqrt{a}+3)-4(\sqrt{a}+3)}{(\sqrt{a}-2)(\sqrt{a}+3)} `

`=\frac{(\sqrt{a}-4)(\sqrt{a}+3)}{(\sqrt{a}-2)(\sqrt{a}+3)} `

`=\frac{\sqrt{a}-4}{\sqrt{a}-2} `

Vậy `P=\frac{\sqrt{a}-4}{\sqrt{a}-2} ` với `a≥0;ane4`

b)

Ta có: `P<1<=>P-1<0`

`=>\frac{\sqrt{a}-4}{\sqrt{a}-2}-1<0`

`<=>\frac{\sqrt{a}-4-(\sqrt{a}-2)}{\sqrt{a}-2}<0`

`<=>\frac{\sqrt{a}-4-\sqrt{a}+2}{\sqrt{a}-2}<0`

`<=>\frac{-2}{\sqrt{a}-2}<0`

Mà `-2<0`

`=>\sqrt{a}-2>0`

`<=>\sqrt{a}>2`

`<=>a>4(N)`

Vậy `a>4` để `P<1`