Đáp án: $x=\dfrac13$
Giải thích các bước giải:
ĐKXĐ: $x\le\dfrac23$
Ta có:
$-3x^2+7x-1=\sqrt{2-3x}$
$\to 3x^2-7x+1+\sqrt{2-3x}=0$
$\to (3x^2-7x+2)+(\sqrt{2-3x}-1)=0$
$\to (3x-1)(x-2)+\dfrac{2-3x-1}{\sqrt{2-3x}+1}=0$
$\to (3x-1)(x-2)+\dfrac{1-3x}{\sqrt{2-3x}+1}=0$
$\to (3x-1)(x-2)-\dfrac{3x-1}{\sqrt{2-3x}+1}=0$
$\to (3x-1)(x-2-\dfrac{1}{\sqrt{2-3x}+1})=0$
$\to 3x-1=0\to x=\dfrac13$
Hoặc $x-2-\dfrac{1}{\sqrt{2-3x}+1}=0$
Mà $x\le \dfrac23$
$\to x-2\le \dfrac23-2<0$
$\to x-2-\dfrac{1}{\sqrt{2-3x}+1}<0$
$\to x-2-\dfrac{1}{\sqrt{2-3x}+1}=0$ vô nghiệm
