`a)` Rút gọn
`ĐK:{(xge0),(xne1/9):}`
`P=((\sqrt{x}-1)/(3\sqrt{x}-1)-1/(3\sqrt{x}+1)+(8\sqrt{x})/(9x-1)):(1-(3\sqrt{x}-2)/(3\sqrt{x}+1))`
`=(((\sqrt{x}-1)(3\sqrt{x}+1)-(3\sqrt{x}-1))/((3\sqrt{x}-1)(3\sqrt{x}+1))+(8\sqrt{x})/((3\sqrt{x}-1)(3\sqrt{x}+1))):(3\sqrt{x}+1-3\sqrt{x}+2)/(3\sqrt{x}+1)`
`=(3x-2\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x})/((3\sqrt{x}-1)(3\sqrt{x}+1)).(3\sqrt{x}+1)/3` `=(3x+3\sqrt{x})/(3(3\sqrt{x}-1))`
`=(3\sqrt{x}(\sqrt{x}+1))/(3(3\sqrt{x}-1))`
`=(x+\sqrt{x})/(3\sqrt{x}-1)`
`b)` `P=6/5`
`->(x+\sqrt{x})/(3\sqrt{x}-1)=6/5`
`->5(x+\sqrt{x})=6(3\sqrt{x}-1)`
`->5x+5\sqrt{x}=18\sqrt{x}-6`
`->5x-13\sqrt{x}+6=0`
`->(5\sqrt{x}-3)(\sqrt{x}-2)=0`
`->`\(\left[ \begin{array}{l}5\sqrt{x}=3\\\sqrt{x}=2\end{array} \right.\)
`->`\(\left[ \begin{array}{l}\sqrt{x}=\frac{3}{5}\\x=2^2\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x=\frac{9}{25}\\x=4\end{array} \right.\)
Vậy ` x={9/25;4}` thì `P=6/5`
