Chứng minh các đẳng thức sau:
a) \(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}=-1,5\)
K
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20 tháng 8 2021
`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`
23 tháng 5 2022
\(VT=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-\dfrac{6\sqrt{6}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{\sqrt{6}}{2}-2\sqrt{6}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{1}{2}-2=-\dfrac{3}{2}\)
2 tháng 8 2017
\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{\sqrt{2^2.3}-\sqrt{6}}{2\sqrt{2}-2}-2\sqrt{6}\right).\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2-1}\right)}-2\sqrt{6}\right).\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{\sqrt{6}}{2}-2\sqrt{6}\right).\dfrac{1}{\sqrt{6}}=\sqrt{6}\left(\sqrt{\dfrac{1}{2}}-2\right).\dfrac{1}{\sqrt{6}}=\dfrac{1}{2}-2=\dfrac{-3}{2}=VP\left(đpcm\right)\)
17 tháng 5 2021
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7 tháng 7 2021
\(=\left(\dfrac{\sqrt{10}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\dfrac{\sqrt{6^2}}{\sqrt{6}}\right)\sqrt{4+\sqrt{15}}\)
\(=\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4+\sqrt{15}}\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{5+2\sqrt{3}\sqrt{5}+3}\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)=5-3=2\)
RK
7 tháng 7 2021
\(VT\Leftrightarrow\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4+\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8+2\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)=5-3=2=VP\left(dpcm\right)\)
a: Ta có: \(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{\sqrt{6}}{2}-\dfrac{4\sqrt{6}}{2}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{-3}{2}\)