Đáp án: $x=\dfrac65$
Giải thích các bước giải:
ĐKXĐ: $0\le x\le 2$
Ta có:
$\sqrt{x(2-x)}+\sqrt{x(3-x)}=\sqrt{6}$
$\to (\sqrt{x(2-x)}+\sqrt{x(3-x)})^2=(\sqrt{6})^2$
$\to x(2-x)+2\sqrt{x(2-x)}\cdot \sqrt{x(3-x)}+x(3-x)=6$
$\to -2x^2+5x+2x\sqrt{(2-x)(3-x)}=6$
$\to 2x\sqrt{(2-x)(3-x)}=2x^2-5x+6$
$\to 2x\sqrt{x^2-5x+6}=2x^2-5x+6$
$\to x^2-2x\sqrt{x^2-5x+6}+(x^2-5x+6)=0$
$\to (x-\sqrt{x^2-5x+6})^2=0$
$\to x-\sqrt{x^2-5x+6}=0$
$\to x=\sqrt{x^2-5x+6}$
$\to x^2=x^2-5x+6$
$\to 5x=6$
$\to x=\dfrac65$
