Câu hỏi của Trần Anh Đức

Question

$\left(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}\right)\left(3\sqrt{\frac{2}{3}}-\sqrt{12}-\sqrt{6}\right)=-\sqrt{2}$

Nguyễn Linh Chi · Accepted Answer

VT$=\left(\frac{3}{2}\sqrt{6}+\frac{2}{3}\sqrt{6}-\frac{4}{2}\sqrt{6}\right)\left(\frac{3}{3}\sqrt{6}-\sqrt{12}-\sqrt{6}\right)$

$=\sqrt{6}\left(\frac{3}{2}+\frac{2}{3}-2\right)\left(-\sqrt{12}\right)$

$=-\sqrt{6}.\frac{1}{6}\sqrt{6.2}=-6.\frac{1}{6}.\sqrt{2}=-\sqrt{2}$= VT

Câu trả lời:

VT$=\left(\frac{3}{2}\sqrt{6}+\frac{2}{3}\sqrt{6}-\frac{4}{2}\sqrt{6}\right)\left(\frac{3}{3}\sqrt{6}-\sqrt{12}-\sqrt{6}\right)$

$=\sqrt{6}\left(\frac{3}{2}+\frac{2}{3}-2\right)\left(-\sqrt{12}\right)$

$=-\sqrt{6}.\frac{1}{6}\sqrt{6.2}=-6.\frac{1}{6}.\sqrt{2}=-\sqrt{2}$= VT