`a)` `A=({2\sqrt{x}}/{\sqrt{x}+3}+\sqrt{x}/{\sqrt{x}-3}-{3x+3}/{x-9}):({2\sqrt{x}-2}/{\sqrt{x}-3}-1)`
$ĐKXĐ: \begin{cases}x\ge 0\\\sqrt{x}+3\ne 0\ (đúng\ với\ mọi\ x\ge 0)\\\sqrt{x}-3\ne 0\\x-9\ne 0\end{cases}$
`<=>`$\begin{cases}x\ge 0\\x\ne 9\end{cases}$
Vậy $ĐKXĐ: x\ge 0;x\ne 9$
$\\$
`b)` `A=({2\sqrt{x}}/{\sqrt{x}+3}+\sqrt{x}/{\sqrt{x}-3}-{3x+3}/{x-9}):({2\sqrt{x}-2}/{\sqrt{x}-3}-1)`
`={2\sqrt{x}(\sqrt{x}-3)+\sqrt{x}(\sqrt{x}+3)-(3x+3)}/{(\sqrt{x}+3)(\sqrt{x}-3)}:{2\sqrt{x}-2-(\sqrt{x}-3)}/{\sqrt{x}-3}`
`={2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}/{(\sqrt{x}+3)(\sqrt{x}-3)}:{\sqrt{x}+1}/{\sqrt{x}-3}`
`={-3\sqrt{x}-3}/{(\sqrt{x}+3)(\sqrt{x}-3)}.{\sqrt{x}-3}/{\sqrt{x}+1}`
`={-3(\sqrt{x}+1)}/{\sqrt{x}+3} . 1/{\sqrt{x}+1}`
`={-3}/{\sqrt{x}+3}`
Vậy: `A={-3}/{\sqrt{x}+3}` với `x\ge 0; x\ne 9`
$\\$
`c)` `A\le -1/3`
`<=>{-3}/{\sqrt{x}+3}\le -1/3`
`<=>{-3}/{\sqrt{x}+3}+1/3\le 0`
`<=>{-3.3+\sqrt{x}+3}/{3(\sqrt{x}+3)}\le 0`
`<=>{\sqrt{x}-6}/{3(\sqrt{x}+3)}\le 0`
`<=>\sqrt{x}-6\le 0` (vì `3(\sqrt{x}+3)>0` với mọi `x\ge 0)`
`<=>\sqrt{x}\le 6`
`<=>(\sqrt{x})^2\le 6^2`
`<=>x\le 36`
Kết hợp điều kiện `x\ge 0;x\ne 9`
`=>0\le x\le 36; x\ne 9` thì `A\le -1/3`
$\\$
`d)` Với mọi `x\ge 0=>\sqrt{x}\ge 0`
`=>\sqrt{x}+3\ge 3`
`=>3/{\sqrt{x}+3}\le 3/3=1`
`=>{-3}/{\sqrt{x}+3}\ge -1`
Dấu "=" xảy ra khi `x=0`
Vậy $GTNN$ của $A$ bằng $-1$ khi $x=0$
