[tex]\left(\frac{\sqrt{98}}{\sqrt{108}}-\frac{\sqrt{324}}{\sqrt{54}}+\frac{\sqrt{48}}{\sqrt{18}}\right)\frac{\sqrt{162}}{\sqrt{363}}[/tex]
[tex]\frac{\sqrt{98}}{\sqrt{108}} = \frac{7\sqrt{2}}{6\sqrt{3}}[/tex]
[tex]\frac{\sqrt{324}}{\sqrt{54}} = \frac{3\sqrt{2}}{\sqrt{3}}[/tex]
[tex]\frac{\sqrt{48}}{\sqrt{18}} = \frac{2\sqrt{6}}{\sqrt{3}}[/tex]
[tex]=\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)[/tex]
[tex]\frac{\sqrt{162}}{\sqrt{363}}= \frac{3\sqrt{6}}{\sqrt{11}}[/tex]
[tex]=\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)[/tex]
[tex]=\frac{1}{3\sqrt{6}}\cdot \frac{3\sqrt{6}}{11}[/tex]
[tex]=\frac{1}{11}[/tex]
b) [tex]\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}}}[/tex]
[tex]\frac{\sqrt{225}}{\sqrt{75}}\cdot \frac{\sqrt{2}}{\sqrt{54}} = \frac{1}{3}[/tex]
[tex]\frac{\frac{\sqrt{24}}{\sqrt{72}}}{\sqrt{\frac{80}{60}}} = \frac{1}{2}[/tex]
[tex]\frac{\sqrt{\frac{147}{75}}}{\frac{42}{\sqrt{25}}} = \frac{1}{6}[/tex]
[tex]=\frac{1}{3}+\frac{1}{2}+\frac{1}{6}[/tex]
[tex]=\frac{2}{6}+\frac{3}{6}+\frac{1}{6}[/tex]
[tex]=\frac{2+3+1}{6}[/tex]
[tex]=\frac{6}{6}[/tex]
[tex]= 1[/tex]
Sper că te-am ajutat!
Succes!
