$\sqrt{3-x}-\sqrt{27-9x}+1,25\sqrt{48-6x}=6_{}$ Tìm x, biết:

$\sqrt{3-x}-\sqrt{27-9x}+1,25\sqrt{48-6x}=6$ $DK:x\leq3$

$⇔\sqrt{3-x}-3\sqrt{3-x}+1,25.4\sqrt{3-x}=6$

$⇔3\sqrt{3-x}=6$

$⇔\sqrt{3-x}=2$

$⇔3-x=4$

$⇔x=-1(tm)$