Đáp án: $x\in\{2-\sqrt{2},1\}$
Giải thích các bước giải:
ĐKXĐ: $x\ge \dfrac12$
Ta có: $\sqrt{2x-1}+x^2-3x+1=0$
$\to x^2-x=2x-1-\sqrt{2x-1}$
$\to 4x^2-4x=4(2x-1)-4\sqrt{2x-1}$
$\to 4x^2-4x+1=4(2x-1)-4\sqrt{2x-1}+1$
$\to (2x-1)^2=(2\sqrt{2x-1}-1)^2$
$\to 2x-1=2\sqrt{2x-1}-1$
$\to 2x=2\sqrt{2x-1}$
$\to x=\sqrt{2x-1}$
$\to x^2=2x-1$
$\to x^2-2x+1=0$
$\to (x-1)^2=0$
$\to x-1=0$
$\to x=1$
Hoặc $2x-1=-2\sqrt{2x-1}+1$
$\to 2x-2=-2\sqrt{2x-1}$
$\to x-1=-\sqrt{2x-1}$
$\to (x-1)^2=(-\sqrt{2x-1})^2$
$\to x^2-2x+1=2x-1$
$\to x^2-4x+2=0$
$\to x=2-\sqrt{2}$ vì $x\ge\dfrac12$
