Question

Exercitiul nr 1 .Efectuati​ dau coroana

Answer 1

$\left(\frac{\sqrt{98}}{\sqrt{108}}-\frac{\sqrt{324}}{\sqrt{54}}+\frac{\sqrt{48}}{\sqrt{18}}\right)\frac{\sqrt{162}}{\sqrt{363}}$

$\frac{\sqrt{98}}{\sqrt{108}} = \frac{7\sqrt{2}}{6\sqrt{3}}$

$\frac{\sqrt{324}}{\sqrt{54}} = \frac{3\sqrt{2}}{\sqrt{3}}$

$\frac{\sqrt{48}}{\sqrt{18}} = \frac{2\sqrt{6}}{\sqrt{3}}$

$=\frac{\sqrt{162}}{\sqrt{363}}\left(\frac{2\sqrt{6}}{3}+\frac{7\sqrt{2}}{6\sqrt{3}}-\frac{3\sqrt{2}}{\sqrt{3}}\right)$

$\frac{\sqrt{162}}{\sqrt{363}}= \frac{3\sqrt{6}}{\sqrt{11}}$

$=\frac{3\sqrt{6}}{11}\left(\frac{2\sqrt{6}}{3}+\frac{7\sqrt{2}}{6\sqrt{3}}-\frac{3\sqrt{2}}{\sqrt{3}}\right)$

$=\frac{1}{3\sqrt{6}}\cdot \frac{3\sqrt{6}}{11}$

$=\frac{1}{11}$

b) $\frac{\sqrt{225}}{\sqrt{75}}\cdot \frac{\sqrt{2}}{\sqrt{54}}+\frac{\frac{\sqrt{24}}{\sqrt{72}}}{\sqrt{\frac{80}{60}}}+\frac{\sqrt{\frac{147}{75}}}{\frac{42}{\sqrt{25}}}$

$\frac{\sqrt{225}}{\sqrt{75}}\cdot \frac{\sqrt{2}}{\sqrt{54}} = \frac{1}{3}$

$\frac{\frac{\sqrt{24}}{\sqrt{72}}}{\sqrt{\frac{80}{60}}} = \frac{1}{2}$

$\frac{\sqrt{\frac{147}{75}}}{\frac{42}{\sqrt{25}}} = \frac{1}{6}$

$=\frac{1}{3}+\frac{1}{2}+\frac{1}{6}$

$=\frac{2}{6}+\frac{3}{6}+\frac{1}{6}$

$=\frac{2+3+1}{6}$

$=\frac{6}{6}$

$= 1$

Sper că te-am ajutat!

Succes!