ĐK: $2\leq x\leq \frac{10}{3}$Ta có: $\sqrt{4-3\sqrt{10-3x}}=x-2\Leftrightarrow 4-3\sqrt{10-3x}=x^2-4x+4$
$\Leftrightarrow x^2-4x+3\sqrt{10-3x}=0\Leftrightarrow (x-3)(x-1)+\frac{9(3-x)}{\sqrt{10-3x}+1}=0$
$\Leftrightarrow x=3$ hoặc $x-1-\frac{9}{\sqrt{10-3x}+1}=0$
Vì $x-1-\frac{9}{\sqrt{10-3x}+1}<0,\forall 2\leq x\leq \frac{10}{3}$
Nên $x=3. (TM)$