Question

Rezultatu calcului radical 6 (3 radical 3 - 4 radical 2) + radical 24 (5 supra radical 2 - 7 supra radical 3) + radical 50 este?​

Answer 1

Răspuns:

$28\sqrt{3} - 33\sqrt{2}$

Explicație pas cu pas:

$6(3\sqrt{3} - 4\sqrt{2} ) + \sqrt{24} (\frac{5}{\sqrt{2} } - \frac{7}{\sqrt{3} } ) + \sqrt{50}$

$= 18\sqrt{3} - 24\sqrt{2} + \frac{5\sqrt{24} }{\sqrt{2} } - \frac{7\sqrt{24} }{\sqrt{3} } + 5\sqrt{2}$

$= 18\sqrt{3} - 24\sqrt{2} + 5\sqrt{12} - 7\sqrt{8} + 5\sqrt{2}$

$= 18\sqrt{3} - 24\sqrt{2} + 10\sqrt{3} - 14\sqrt{2} + 5\sqrt{2}$

$= 28\sqrt{3} - 33\sqrt{2}$

Answer 2

Explicație pas cu pas:

$\sqrt{6}(3 \sqrt{3} - 4 \sqrt{2}) + \sqrt{24} \bigg( \dfrac{5}{ \sqrt{2} } - \dfrac{7}{ \sqrt{3} } \bigg) + \sqrt{50} =$

$= 3 \sqrt{18} - 4 \sqrt{12} + 2\sqrt{6} \bigg( \dfrac{5}{ \sqrt{2} } - \dfrac{7}{ \sqrt{3} } \bigg) + 5\sqrt{2}$

$= 3 \cdot 3 \sqrt{2} - 4 \cdot 2 \sqrt{3} + \dfrac{10\sqrt{6}}{ \sqrt{2} } - \dfrac{14\sqrt{6}}{ \sqrt{3} } + 5\sqrt{2}$

$= 9 \sqrt{2} - 8 \sqrt{3} + 10 \sqrt{3} - 14 \sqrt{2} + 5\sqrt{2}$

$= 14 \sqrt{2} - 14 \sqrt{2} + 2 \sqrt{3} = \bf 2 \sqrt{3}$