Giải thích các bước giải:
DKXD : $2\le x\le \dfrac{10}{3}$
$\sqrt{x-2}+\sqrt{10-3x}=5-x$
$\rightarrow \sqrt{x-2}+\sqrt{10-3x}=\dfrac{1}{2}(x-2+10-3x+2)$
$\rightarrow 2\sqrt{x-2}+2\sqrt{10-3x}=(x-2)+(10-3x)+2$
$\rightarrow((x-2)- 2\sqrt{x-2}+1)+((10-3x)-2\sqrt{10-3x}+1)^2=0$
$\rightarrow(\sqrt{x-2}-1)^2+(\sqrt{10-3x}-1)^2=0$
$\rightarrow(\sqrt{x-2}-1)^2=(\sqrt{10-3x}-1)^2=0$
$\rightarrow x=3$
