Question

28-b plssssssss
va roggg

Answer 1

Explicație pas cu pas:

$\dfrac{ \sqrt{2} }{2} + { \bigg( \dfrac{4 \sqrt{3} - 3 \sqrt{2} }{6} \bigg)}^{2} \cdot \dfrac{4 \sqrt{3} + 3 \sqrt{2} }{5} =$

$= \dfrac{ \sqrt{2} }{2} + \dfrac{(4 \sqrt{3} - 3 \sqrt{2})(4 \sqrt{3} - 3 \sqrt{2})(4 \sqrt{3} + 3 \sqrt{2}) }{ {6}^{2} \cdot 5}$

$= \dfrac{ \sqrt{2} }{2} + \dfrac{(4 \sqrt{3} - 3 \sqrt{2})(48 - 18)}{6 \cdot 30} = \dfrac{ \sqrt{2} }{2} + \dfrac{(4 \sqrt{3} - 3 \sqrt{2}) \cdot 30}{6 \cdot 30}$

$= \dfrac{3\sqrt{2} + 4 \sqrt{3} - 3 \sqrt{2}}{6} = \dfrac{4 \sqrt{3}}{6} = \bf \dfrac{2 \sqrt{3}}{3}$