$\sqrt{3-x}-\sqrt{27-9x}+1,25\sqrt{48-6x}=6_{}$ Tìm x, biết: 23/07/2021 Bởi Sarah $\sqrt{3-x}-\sqrt{27-9x}+1,25\sqrt{48-6x}=6_{}$ Tìm x, biết:
$\sqrt{3-x}-\sqrt{27-9x}+1,25\sqrt{48-6x}=6$ $DK:x\leq3$ $⇔\sqrt{3-x}-3\sqrt{3-x}+1,25.4\sqrt{3-x}=6$ $⇔3\sqrt{3-x}=6$ $⇔\sqrt{3-x}=2$ $⇔3-x=4$ $⇔x=-1(tm)$ Bình luận
$\sqrt{3-x}-\sqrt{27-9x}+1,25\sqrt{48-6x}=6$ $DK:x\leq3$
$⇔\sqrt{3-x}-3\sqrt{3-x}+1,25.4\sqrt{3-x}=6$
$⇔3\sqrt{3-x}=6$
$⇔\sqrt{3-x}=2$
$⇔3-x=4$
$⇔x=-1(tm)$