Question

Решите задания во вкладках.........

Answer 1

$\frac{1}{3} (2 \sqrt{150} +3 \sqrt{24} -5 \sqrt{54} )^2+15\cdot \sqrt[4]{625} = \\\ = \frac{1}{3} (2 \sqrt{25\cdot6} +3 \sqrt{4\cdot6} -5 \sqrt{9\cdot6} )^2+15\cdot \sqrt[4]{5^4} = \\\ = \frac{1}{3} (10 \sqrt{6} +6 \sqrt{6} -15 \sqrt{6} )^2+15\cdot 5 = \frac{1}{3} (\sqrt{6})^2+75= \\\ =\frac{1}{3} \cdot6+75=2+75=77$

$\sqrt{20.25} + \sqrt[3]{24} - \sqrt[4]{0.1296} - \frac{2}{5} \sqrt[3]{375} + \frac{1}{3} \sqrt[5]{7 \frac{19}{32} } = \\\ =\sqrt{4.5^2} + \sqrt[3]{8\cdot3} - \sqrt[4]{0.6^4} - \frac{2}{5} \sqrt[3]{125\cdot3} + \frac{1}{3} \sqrt[5]{\frac{243}{32} } = \\\ = 4.5 + 2\sqrt[3]{3} -0.6 - \frac{2}{5}\cdot5 \sqrt[3]{3} + \frac{1}{3} \sqrt[5]{(\frac{3}{2})^5 } \\\ =3.9 + 2\sqrt[3]{3} - 2 \sqrt[3]{3} + \frac{1}{3}\cdot\frac{3}{2}= 3.9 + \frac{1}{2} =3.9+0.5=4.4$

Answer 2

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