tìm x (2 √x +1)( √x-2)=7

(2$\sqrt{x}$ `+1)(`     $\sqrt{x}$`-2)`    `=7`

`ĐK : x≥0`

`⇔2x -4`$\sqrt{x}$ `+` $\sqrt{x}$ `-2 = 7`

`⇔2x - 3`$\sqrt{x}$ `- 9 = 0`

`⇔`($\sqrt{x}$`- 3)(`$\sqrt{x}$ `+ 3/2)`   `=0`

`⇔x = 9(TM)` hoặc $\sqrt{x}$`= -3/2(vô lí)`

(2$\sqrt{x}$ + 1)($\sqrt{x}$ - 2) = 7 Đkxđ: x ≥ 0

⇔ 2x - 4$\sqrt{x}$ + $\sqrt{x}$ - 2 = 7

⇔ 2x - 3$\sqrt{x}$ - 9 = 0

⇔ 2x - 6$\sqrt{x}$ + 3$\sqrt{x}$ - 9 = 0

⇔ 2$\sqrt{x}$($\sqrt{x}$ - 3) + 3($\sqrt{x}$ - 3) = 0

⇔ ($\sqrt{x}$ - 3)(2$\sqrt{x}$ + 3) = 0

⇔ \(\left[ \begin{array}{l}\sqrt{x} - 3 = 0\\2\sqrt{x} + 3 = 0\end{array} \right.\) 

⇔ \(\left[ \begin{array}{l}x = 9 (t/m)\\x ∈ ∅\end{array} \right.\) 

 Vậy x = 9

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