Đáp án:
`a)` `P={4\sqrt{x}}/{3(x-\sqrt{x}+1)}` với `x\ge 0`
`b)` `x\in {1/4;4}`
Giải thích các bước giải:
`a)` `P=({x+2}/{x\sqrt{x}+1}-1/{\sqrt{x}+1}).{4\sqrt{x}}/3\quad (x\ge 0)`
`=({x+2}/{(\sqrt{x})^3+1^3}-1/{\sqrt{x}+1}).{4\sqrt{x}}/3`
`={x+2-(x-\sqrt{x}+1)}/{(\sqrt{x}+1)(x-\sqrt{x}+1)}. {4\sqrt{x}}/3`
`={\sqrt{x}+1}/{(\sqrt{x}+1)(x-\sqrt{x}+1)}. {4\sqrt{x}}/3`
`={4\sqrt{x}}/{3(x-\sqrt{x}+1)}`
Vậy `P={4\sqrt{x}}/{3(x-\sqrt{x}+1)}` với `x\ge 0`
$\\$
`b)` Để `P=8/9`
`<=>{4\sqrt{x}}/{3(x-\sqrt{x}+1)}=8/9`
`<=>9.4\sqrt{x}=8.3(x-\sqrt{x}+1)`
`<=>36\sqrt{x}=24x-24\sqrt{x}+24`
`<=>24x-60\sqrt{x}+24=0`
`<=>2x-5\sqrt{x}+2=0` $(1)$
Đặt `t=\sqrt{x}\quad (t\ge 0)`
`(1)<=>2t^2-5t+2=0`
`<=>`$\left[\begin{array}{l}t=\dfrac{1}{2}\\t=2\end{array}\right.$(thỏa mãn)
`<=>`$\left[\begin{array}{l}\sqrt{x}=\dfrac{1}{2}\\\sqrt{x}=2\end{array}\right.$
`<=>`$\left[\begin{array}{l}x=\dfrac{1}{4}\ \\x=4\end{array}\right.$ (thỏa mãn)
Vậy `x\in {1/4;4}` thì `P=8/9`
