Câu hỏi của Nguyễn Tấn An

Question

Chứng minh: ($\left(\sqrt{12}-6\sqrt{3}+\sqrt{24}\right)\sqrt{6}-\left(5\sqrt{\dfrac{1}{12}}+12\right)=-14,5\sqrt{2}$

Nhã Doanh · Accepted Answer

$\left(\sqrt{12}-6\sqrt{3}+\sqrt{24}\right)\sqrt{6}-\left(5\sqrt{\dfrac{1}{12}}+12\right)$

$=\left(2\sqrt{3}-6\sqrt{3}+2\sqrt{6}\right)\sqrt{6}-\left(\dfrac{5\sqrt{3}}{6}+12\right)$

$=6\sqrt{2}-18\sqrt{2}+12-\left(\dfrac{5\sqrt{3}+72}{6}\right)$

$=-12\sqrt{2}+12-\dfrac{5\sqrt{3}+72}{6}$

$=\dfrac{-72\sqrt{2}+72-5\sqrt{3}-72}{6}=\dfrac{5\sqrt{3}+72\sqrt{2}}{6}\simeq-18,4139$

Ta có: $-14,5\sqrt{2}\simeq-20,506$

$VT\ne VP$

Đẳng thức không xảy raCâu trả lời: $\left(\sqrt{12}-6\sqrt{3}+\sqrt{24}\right)\sqrt{6}-\left(5\sqrt{\dfrac{1}{12}}+12\right)$

$=\left(2\sqrt{3}-6\sqrt{3}+2\sqrt{6}\right)\sqrt{6}-\left(\dfrac{5\sqrt{3}}{6}+12\right)$

$=6\sqrt{2}-18\sqrt{2}+12-\left(\dfrac{5\sqrt{3}+72}{6}\right)$

$=-12\sqrt{2}+12-\dfrac{5\sqrt{3}+72}{6}$

$=\dfrac{-72\sqrt{2}+72-5\sqrt{3}-72}{6}=\dfrac{5\sqrt{3}+72\sqrt{2}}{6}\simeq-18,4139$

Ta có: $-14,5\sqrt{2}\simeq-20,506$

$VT\ne VP$

Đẳng thức không xảy ra