a. Tử `=1-(a-3\sqrt{a})/(a-9)=1-(\sqrt{a}(\sqrt{a}-3))/((\sqrt{a}-3)(\sqrt{a}+3))`
`=1-(\sqrt{a})/(\sqrt{a}+3)=3/(\sqrt{a}+3)`
Mẫu `=(\sqrt{a}-2)/(\sqrt{a}+3)+(\sqrt{a}-3)/(2-\sqrt{a})-(9-a)/(a+\sqrt{a}-6)`
`=(\sqrt{a}-2)/(\sqrt{a}+3)-(\sqrt{a}-3)/(\sqrt{a}-2)+(a-9)/((\sqrt{a}-2)(\sqrt{a}+3))`
`=(\sqrt{a}-2)/(\sqrt{a}+3)-(\sqrt{a}-3)/(\sqrt{a}-2)+(\sqrt{a}-3)/(\sqrt{a}-2)`
`=(\sqrt{a}-2)/(\sqrt{a}+3)`
`⇒A=3/(\sqrt{a}+3):(\sqrt{a}-2)/(\sqrt{a}+3)=3/(\sqrt{a}-2)`
Vậy `A=3/(\sqrt{a}-2)` khi $a\neq9,a\neq4,a≥0$
b. `A-1/A=0` $Đk:a\neq9,a\neq4,a≥0$
`⇒3/(\sqrt{a}-2)-(\sqrt{a}-2)/3=0`
`⇔9-a+4\sqrt{a}-4=0`
`⇔a-4\sqrt{a}-5=0`
`⇔(\sqrt{a}-5)(\sqrt{a}+1)=0`
$⇔\left[ \begin{array}{l}a=25\ (tm)\\\sqrt{a}=-1\ (loại)\end{array} \right.$
Vậy $a=25$ thì `A-1/A=0`.
