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a: Ta có: \(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{\sqrt{6}}{2}-\dfrac{4\sqrt{6}}{2}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{-3}{2}\)
\(VT=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-\dfrac{6\sqrt{6}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{\sqrt{6}}{2}-2\sqrt{6}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{1}{2}-2=-\dfrac{3}{2}\)
\(=\left(2\sqrt{3}-2\sqrt{6}\right)\cdot\sqrt{6}=6\sqrt{2}-12\)
\(B=\sqrt{9+4\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)
\(B=\sqrt{\left(\sqrt{5}+2\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(B=\left|\sqrt{5}+2\right|+\left|\sqrt{5}-2\right|\)
\(B=\sqrt{5}+2+\sqrt{5}-2\)
\(B=2\sqrt{5}\)
\(A=\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}\)
\(A=\left(\dfrac{\sqrt{12}-\sqrt{6}}{2\sqrt{2}-2}-\dfrac{6\sqrt{6}}{3}\right).\dfrac{1}{\sqrt{6}}\)
\(A=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right).\dfrac{1}{\sqrt{6}}\)
\(A=\left(\sqrt{6}-2\sqrt{6}\right).\dfrac{1}{\sqrt{6}}\)
\(A=-\sqrt{6}.\dfrac{1}{\sqrt{6}}\)
\(A=-1\)
`e)(3/2sqrt6+2sqrt{2/3}-4sqrt{3/2})(3/2sqrt6+2sqrt{2/3}+4sqrt{3/2})`
`=(3/2sqrt6+2sqrt{2/3})^2-(4\sqrt{3/2})^2`
`=((3sqrt6)/2+(2sqrt2)/3)^2-16*3/2`
`=((9sqrt6)/6+(4sqrt6)/6)^2-24`
`=((13sqrt6)/6)^2-24`
`=13^2/6-24`
`=25/6`
a: \(=\dfrac{6}{4+\sqrt{3}-1}=\dfrac{6}{3+\sqrt{3}}=3-\sqrt{3}\)
b: \(=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right)\cdot\dfrac{1}{\sqrt{6}}=\sqrt{6}\cdot\dfrac{1}{\sqrt{6}}\left(\dfrac{1}{2}-2\right)=-\dfrac{3}{2}\)
\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{\sqrt{2^2.3}-\sqrt{6}}{2\sqrt{2}-2}-2\sqrt{6}\right).\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2-1}\right)}-2\sqrt{6}\right).\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{\sqrt{6}}{2}-2\sqrt{6}\right).\dfrac{1}{\sqrt{6}}=\sqrt{6}\left(\sqrt{\dfrac{1}{2}}-2\right).\dfrac{1}{\sqrt{6}}=\dfrac{1}{2}-2=\dfrac{-3}{2}=VP\left(đpcm\right)\)
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