Question

Să se calculeze sumele:

Answer 1

Răspuns:

$\displaystyle -\frac{7}{6}\sqrt[4]{\frac{1}{2}}$

Rezolvare:

$\displaystyle 2\sqrt[4]{\frac{1}{2}}-\sqrt[4]{\frac{81}{32}}-\sqrt[4]{\frac{625}{162}} =$

$\displaystyle =2\sqrt[4]{\frac{1}{2}}-\sqrt[4]{\frac{3^4}{2\cdot 16}}-\sqrt[4]{\frac{5^4}{2\cdot 81}}$

$\displaystyle = 2\sqrt[4]{\frac{1}{2}}-\sqrt[4]{\frac{3^4}{2\cdot 2^4}}-\sqrt[4]{\frac{5^4}{2\cdot 3^4}}$

$\displaystyle = 2\sqrt[4]{\frac{1}{2}}-\sqrt[4]{\frac{1}{2}\cdot \frac{3^4}{2^4}}-\sqrt[4]{\frac{1}{2}\cdot \frac{5^4}{3^4}}$

$\displaystyle = 2\sqrt[4]{\frac{1}{2}}-\sqrt[4]{\frac{1}{2}}\cdot \sqrt[4]{\frac{3^4}{2^4}}-\sqrt[4]{\frac{1}{2}}\cdot \sqrt[4]{\frac{5^4}{3^4}}$

$\displaystyle = 2\sqrt[4]{\frac{1}{2}}-\frac{3}{2}\sqrt[4]{\frac{1}{2}}-\frac{5}{3}\sqrt[4]{\frac{1}{2}}$

$\displaystyle = \sqrt[4]{\frac{1}{2}}\cdot \left(^{^{^{\displaystyle 6)}}}\!\!2-^{^{^{\displaystyle 3)}}}\!\!\frac{3}{2}-^{^{^{\displaystyle 2)}}}\!\!\frac{5}{3}\right)$

$\displaystyle = \sqrt[4]{\frac{1}{2}}\cdot \left(\frac{2\cdot 6}{6}-\frac{3\cdot 3}{6}-\frac{5\cdot 2}{6}\right)$

$= \displaystyle \sqrt[4]{\frac{1}{2}}\cdot \left(\frac{12-9-10}{6}\right)$

$\displaystyle = \sqrt[4]{\frac{1}{2}}\cdot \left(-\frac{7}{6}\right)$

$\displaystyle = \boxed{-\frac{7}{6}\sqrt[4]{\frac{1}{2}}}$