ta tính VT ra xong rồi nói VT = VP

a)\(\frac{3.\sqrt{6}}{2}+\frac{2.\sqrt{2}}{\sqrt{3}}-\frac{4.\sqrt{3}}{\sqrt{2}}=\frac{3\sqrt{6}}{2}+\frac{2\sqrt{2}.\sqrt{3}}{\sqrt{3}.\sqrt{3}}-\frac{4.\sqrt{3}.\sqrt{2}}{\sqrt{2}.\sqrt{2}}=\frac{3\sqrt{6}}{2}+\frac{2\sqrt{6}}{3}-\frac{4\sqrt{6}}{2}=\frac{2\sqrt{6}}{3}-\frac{\sqrt{6}}{2}=\frac{4\sqrt{6}-3\sqrt{6}}{6}=\frac{\sqrt{6}}{6}\)

--> dpcm

b) \(\left(\frac{-\sqrt{7}.\left(1-\sqrt{2}\right)}{1-\sqrt{2}}+\frac{-\sqrt{5}.\left(1-\sqrt{3}\right)}{1-\sqrt{3}}\right).\frac{\sqrt{7}-\sqrt{5}}{1}\)

=\(\left(-\sqrt{7}-\sqrt{5}\right).\left(\sqrt{7}-\sqrt{5}\right)\)

=\(-1.\left(\sqrt{7}+\sqrt{5}\right).\left(\sqrt{7}-\sqrt{5}\right)\)

=\(-1.\left(7-5\right)\)

=-1.2

=-2

Biến đổi Vế trái ta có :

 \(\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{3}\left(2-\sqrt{2}\right)}{\sqrt{2}\left(2-\sqrt{2}\right)}-\frac{3\sqrt{24}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)

\(\left(\frac{\sqrt{3}}{\sqrt{2}}-2\sqrt{6}\right)\cdot\frac{1}{\sqrt{6}}=\frac{\sqrt{3}}{\sqrt{2}}\cdot\frac{1}{\sqrt{6}}-2\sqrt{6}\cdot\frac{1}{\sqrt{6}}\)

\(=\frac{1}{2}-2=-1,5=VP\)  ( ĐPCM) 

Biến đổi vế trái :

\(VT=\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right).\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{2}.\sqrt{2}.\sqrt{3}-\sqrt{6}}{\sqrt{2^2.2}-2}-\frac{\sqrt{6^2.6}}{3}\right).\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{2}.\sqrt{6}-\sqrt{6}}{2\sqrt{2}-2}-\frac{6\sqrt{6}}{3}\right).\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{6}.\left(\sqrt{2}-1\right)}{2.\left(\sqrt{2}-1\right)}-2\sqrt{6}\right).\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{6}}{2}-2\sqrt{6}\right).\frac{1}{\sqrt{6}}\)

\(=\sqrt{6}.\left(\frac{1}{2}-2\right).\frac{1}{\sqrt{6}}=-\frac{3}{2}=-1,5=VP\left(đpcm\right)\)

a) Biến đổi vế trái ta có:
\(\frac{3}{2}\sqrt{6}+2\sqrt{\frac{2}{3}}-4\sqrt{\frac{3}{2}}\)

\(=\frac{3\sqrt{6}}{2}+\frac{2\sqrt{6}}{3}-\frac{4\sqrt{6}}{2}=\frac{9\sqrt{6}+4\sqrt{6}-12\sqrt{6}}{6}=\frac{\sqrt{6}}{6}=VP\)

Vậy đẳng thức trên đc chứng minh

b) Biến đổi vế trái ta có:

\(\left(x\sqrt{\frac{6}{x}}+\sqrt{\frac{2x}{3}}+\sqrt{6x}\right):\sqrt{6x}\)

\(=\left(x\sqrt{\frac{6}{x}}+\sqrt{\frac{2x}{3}}+\sqrt{6x}\right)\cdot\frac{1}{\sqrt{6x}}\)

\(=x\sqrt{\frac{6}{x}\cdot\frac{1}{6x}}+\sqrt{\frac{2x}{3}\cdot\frac{1}{6x}}+\sqrt{6x}\cdot\frac{1}{\sqrt{6x}}\)

\(=x\sqrt{\frac{1}{x^2}}+\sqrt{\frac{1}{9}}+1=1+\frac{1}{3}+1=2\frac{1}{3}=VP\)

Vậy đẳng thức trên đc chứng minh