Đáp án: $x=2$
Giải thích các bước giải:
ĐKXĐ: $x\ge 2$
Ta có:
$\sqrt{x^2-3x+2}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{x^2+2x-3}$
$\to \sqrt{(x-1)(x-2)}+\sqrt{x+3}=\sqrt{x-2}+\sqrt{(x-1)(x+3)}$
$\to \sqrt{(x-1)(x-2)}-\sqrt{x-2}=\sqrt{(x-1)(x+3)}-\sqrt{x+3}$
$\to \sqrt{x-2}(\sqrt{x-1}-1)=\sqrt{x+3}(\sqrt{x-1}-1)$
$\to \sqrt{x-2}(\sqrt{x-1}-1)-\sqrt{x+3}(\sqrt{x-1}-1)=0$
$\to(\sqrt{x-1}-1)(\sqrt{x-2}-\sqrt{x+3})=0$
$\to \sqrt{x-1}-1=0\to \sqrt{x-1}=1\to x-1=1\to x=2$
Hoặc $\sqrt{x-2}-\sqrt{x+3}=0\to \sqrt{x-2}=\sqrt{x+3}\to x-2=x+3\to 0=5$ vô lý
