Question

Modului lui $(- \sqrt{2})^{3}$ + $(- \sqrt{2})^{5}$ - aici se incheie modului impartit la (-2√6) este......

Answer 1

$\displaystyle \left|\left(- \sqrt{2} \right)^3+\left(- \sqrt{2} \right)^5\right|:(-2 \sqrt{6} )= \\ \\ =\left|(- \sqrt{2} ) \cdot (- \sqrt{2} ) \cdot {(- \sqrt{2} )+(- \sqrt{2} ) \cdot (- \sqrt{2} )\cdot (- \sqrt{2} ) \cdot (- \sqrt{2} ) \cdot (- \sqrt{2} )\right|: \\ :(-2 \sqrt{6} )= \\ \\ =\left|-2 \sqrt{2} +2 \cdot 2 \cdot (- \sqrt{2} )\right|:(-2 \sqrt{6} )=\left|-2 \sqrt{2} -4 \sqrt{2} \right|:(-2 \sqrt{6} )=$

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