Question

Va rog ajutorrrrrrr.
$( \frac{4 - \sqrt{3} }{ \sqrt{3} } + \frac{2 \sqrt{3} - 3 }{ \sqrt{12} } ) \times ( \frac{2}{ \sqrt{6} } + \frac{3 \sqrt{6} }{2} - \frac{12 \sqrt{2} }{4 \sqrt{3} } ) =$​

Answer 1

Răspuns:

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Answer 2

Explicație pas cu pas:

$\bigg( \dfrac{4 - \sqrt{3} }{ \sqrt{3} } + \dfrac{2 \sqrt{3} - 3 }{ \sqrt{12} } \bigg) \cdot \bigg( \dfrac{2}{ \sqrt{6} } + \dfrac{3 \sqrt{6} }{2} - \dfrac{12 \sqrt{2} }{4 \sqrt{3} } \bigg) =$

$= \bigg( \dfrac{4 - \sqrt{3} }{ \sqrt{3} } + \dfrac{2 \sqrt{3} - 3 }{ 2\sqrt{3} } \bigg) \cdot \bigg( \dfrac{2}{ \sqrt{6} } + \dfrac{3 \sqrt{6} }{2} - \dfrac{6 \sqrt{2} }{2 \sqrt{3} } \bigg)$

$= \bigg( \dfrac{2(4 - \sqrt{3})}{2\sqrt{3} } + \dfrac{2 \sqrt{3} - 3 }{ 2\sqrt{3} } \bigg) \cdot \bigg( \dfrac{2 \sqrt{6} }{6} + \dfrac{9 \sqrt{6} }{6} - \dfrac{6 \sqrt{3} }{6} \bigg)$

$= \dfrac{8 - 2\sqrt{3} + 2 \sqrt{3} - 3}{2\sqrt{3} } \cdot \dfrac{2 \sqrt{6} + 9 \sqrt{6} - 6 \sqrt{3} }{6}$

$= \dfrac{5}{2\sqrt{3} } \cdot \dfrac{5 \sqrt{6}}{6} = \dfrac{5}{2} \cdot \dfrac{5 \sqrt{2}}{6} = \dfrac{25 \sqrt{2}}{12}$