Đáp án:

 `b)` `Q=-2\sqrt{5}-6` khi `a=2/{3-\sqrt{5}}`

`c)` `0\le a<4`

Giải thích các bước giải:

`Q={\sqrt{a}+3}/{\sqrt{a}-2}-{\sqrt{a}-1}/{\sqrt{a}+2}-{4\sqrt{a}-4}/{a-4}` `(a\ge 0;a\ne 4)`

`={(\sqrt{a}+3)(\sqrt{a}+2)-(\sqrt{a}-1)(\sqrt{a}-2)-(4\sqrt{a}-4)}/{(\sqrt{a}-2)(\sqrt{a}+2)}`

`={a+2\sqrt{a}+3\sqrt{a}+6-(a-2\sqrt{a}-\sqrt{a}+2)-4\sqrt{a}+4}/{(\sqrt{a}-2)(\sqrt{a}+2)}`

`={a+5\sqrt{a}+6-a+3\sqrt{a}-2-4\sqrt{a}+4}/{(\sqrt{a}-2)(\sqrt{a}+2)}`

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

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

$\\$

`b)` `a=2/{3-\sqrt{5}}` (thỏa mãn đk)

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

`={6+2\sqrt{5}}/{9-5}`

`={5+2.\sqrt{5}.1+1^2}/4`

`={(\sqrt{5}+1)^2}/{2^2}=({\sqrt{5}+1}/2)^2`

`=>\sqrt{a}={\sqrt{5}+1}/2`

`=>\sqrt{a}-2={\sqrt{5}+1}/2-2`

`={\sqrt{5}+1-4}/2={\sqrt{5}-3}/2`

Ta có:

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

$=\dfrac{4}{\dfrac{\sqrt{5}-3}{2}}$

`={4.2}/{\sqrt{5}-3}`

`={4.2.(\sqrt{5}+3)}/{(\sqrt{5}-3)(\sqrt{5}+3)}`

`={4.2.(\sqrt{5}+3)}/{5-9}=-2(\sqrt{5}+3)`

`=-2\sqrt{5}-6`

Vậy `Q=-2\sqrt{5}-6` khi `a=2/{3-\sqrt{5}}`

$\\$

`c)` `Q\le 0`

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

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

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

`<=>a<4`

Kết hợp điều kiện `a\ge 0;a\ne 4`

`=>0\le a<4` thì `Q\le 0`