giúp mình câu c

`A={sqrt{x}}/{sqrt{x}+3}+{2sqrt{x}}/{sqrt{x}-3}-{2x}/{x-9}` `(x>=0;x\ne9)`

`={sqrt{x}}/{sqrt{x}+3}+{2sqrt{x}}/{sqrt{x}-3}-{2x}/{(sqrt{x}-3)(sqrt{x}+3)}`

`={sqrt{x}(sqrt{x}-3)+2sqrt{x}(sqrt{x}+3)-2x}/{(sqrt{x}-3)(sqrt{x}+3)}`

`={x-3sqrt{x}+2x+6sqrt{x}-2x}/{(sqrt{x}-3)(sqrt{x}+3)}`

`={x+3sqrt{x}}/{(sqrt{x}-3)(sqrt{x}+3)}`

`={sqrt{x}}/{sqrt{x}-3}`

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`B-8A<=0<=> {x+16}/{sqrt{x}-3}-{8sqrt{x}}/{sqrt{x}-3}<=0`

`<=> {x-8sqrt{x}+16}/{sqrt{x}-3}<=0`

`<=> {(sqrt{x}-4)^2}/{sqrt{x}-3}<=0`

`<=>`\(\left[ \begin{array}{l}\sqrt{x}-4=0\\\sqrt{x}-3<0\end{array} \right.\)

`<=>`\(\left[ \begin{array}{l}x=16\\x<9\end{array} \right.\)

Kết hợp điều kiện : `=>`\(\left[ \begin{array}{l}x=16(nhận)\\ 0\le x<9\end{array} \right.\)

Vậy `x=16` hoặc `0<=x<9` thì thỏa ycbt.