Rút gọn C :
\(C=\dfrac{2}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}+\dfrac{2\sqrt{x}-3}{9-x}\left(dkxd:x\ge0,x\ne9\right)\)
\(=\dfrac{2}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}-\dfrac{2\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{2\left(\sqrt{x}+3\right)+\sqrt{x}-3-\left(2\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{2\sqrt{x}+6+\sqrt{x}-3-2\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{6+\sqrt{x}}{x-9}\)
Để \(C=-\dfrac{8}{5}\) thì \(\dfrac{6+\sqrt{x}}{x-9}=-\dfrac{8}{5}\)
\(\Rightarrow5\left(6+\sqrt{x}\right)=-8\left(x-9\right)\)
\(\Rightarrow30+5\sqrt{x}+8x-72=0\)
\(\Rightarrow5\sqrt{x}+8x-42=0\)
\(\Rightarrow64x=441\)
\(\Rightarrow x=4\)
Vậy \(x=4\) thì \(C=-\dfrac{8}{5}\)
Với `x\ne0;x\ne9` ta có:
`C=2/(sqrtx-3)+1/(sqrtx+3)+(2sqrtx-3)/(9-x)=(2(sqrtx+3)+(sqrtx-3)-(2sqrtx-3))/((sqrtx-3)(sqrtx+3))=(2sqrtx+6+sqrtx-3-2sqrtx+3)/((sqrtx-3)(sqrtx+3))=(sqrtx+6)/(x-9)`
Để `C=-8/5` thì `(sqrtx+6)/(x-9)=-8/5`
`<=>5(sqrtx+6)=-8(x-9)`
`<=>5sqrtx+30=-8x+72`
`<=>8x+5sqrtx-42=0`
`<=>(sqrtx-2)(8sqrtx+21)=0`
`=>sqrtx-2=0``
`<=>sqrtx=2`
`=>x=4(TMĐK)`
Vậy `x=4` thì `C=-8/5`
