Đáp án: $x\in\{1,\dfrac{-3-\sqrt{33}}{4}\}$
Giải thích các bước giải:
ĐKXĐ: $x\ge 1$ hoặc $x\le-2$
Ta có:
$\sqrt{2x^2-3x+1}+\sqrt{x^2+x-2}=\sqrt{3x^2-4x+1}$
$\to (\sqrt{2x^2-3x+1}+\sqrt{x^2+x-2})^2=(\sqrt{3x^2-4x+1})^2$
$\to 2x^2-3x+1+x^2+x-2+2\sqrt{2x^2-3x+1}\cdot\sqrt{x^2+x-2}=3x^2-4x+1$
$\to 3x^2-2x-1+2\sqrt{(2x^2-3x+1)(x^2+x-2)}=3x^2-4x+1$
$\to 2\sqrt{(2x^2-3x+1)(x^2+x-2)}=-2x+2$
$\to \sqrt{(2x-1)(x-1)(x-1)(x+2)}=-x+1$
$\to \sqrt{(2x-1)(x-1)^2(x+2)}=-x+1$
$\to |x-1|\sqrt{(2x-1)(x+2)}=-(x-1)$
Nếu $x\ge 1$
$\to (x-1)\sqrt{(2x-1)(x+2)}=-(x-1)$
$\to (x-1)\sqrt{(2x-1)(x+2)}+(x-1)=0$
$\to (x-1)(\sqrt{(2x-1)(x+2)}+1)=0$
$\to x-1=0$ vì $\sqrt{(2x-1)(x+2)}+1>0$
$\to x>1$
Nếu $x\le -2$
$\to -(x-1)\sqrt{(2x-1)(x+2)}=-(x-1)$
$\to \sqrt{(2x-1)(x+2)}=1$ vì $x\le -2\to x-1\ne 0$
$\to (2x-1)(x+2)=1$
$\to 2x^2+3x-2=1$
$\to 2x^2+3x-3=0$
$\to x=\dfrac{-3-\sqrt{33}}{4}$ vì $x\le -2$
