Question

exercitiul 14 va rooog

Answer 1

   
[tex]\displaystyle\\c)\\ \left( \frac{4}{ \sqrt{32}} +\frac{6}{ \sqrt{72}} +\frac{8}{ \sqrt{128}} +\frac{10}{ \sqrt{200}} \right): \sqrt{8}=\\\\ = \left( \frac{4}{ \sqrt{16\times2}} +\frac{6}{ \sqrt{36\times2}} +\frac{8}{ \sqrt{64\times2}} +\frac{10}{ \sqrt{100\times2}} \right): \sqrt{4\times2}=\\\\ = \left( \frac{4}{ 4\sqrt{2}} +\frac{6}{ 6\sqrt{2}} +\frac{8}{8 \sqrt{2}} +\frac{10}{ 10\sqrt{2}} \right): 2\sqrt{2}= [/tex]

[tex]\displaystyle\\ = \left( \frac{1}{\sqrt{2}} +\frac{1}{\sqrt{2}} +\frac{1}{\sqrt{2}} +\frac{1}{\sqrt{2}} \right): 2\sqrt{2}=\\\\ = \left( \frac{\sqrt{2}}{2} +\frac{\sqrt{2}}{2} +\frac{\sqrt{2}}{2} +\frac{\sqrt{2}}{2} \right): 2\sqrt{2}=\\\\ =\frac{4\sqrt{2}}{2}: 2\sqrt{2}= 2\sqrt{2}:2\sqrt{2}=\boxed{\bf 1}[/tex]



[tex]\displaystyle\\ d)\\ \left(\frac{\sqrt{21}+\sqrt{14}}{\sqrt{6}+2}+\frac{\sqrt{35}+\sqrt{21}}{\sqrt{10}+\sqrt{6}}\right):\frac{7}{\sqrt{14}}=\\\\\\ =\left(\frac{\sqrt{3\times7}+\sqrt{2\times7}}{\sqrt{3\times2}+ \sqrt{2\times 2}}+\frac{\sqrt{5\times7}+\sqrt{3\times7}}{\sqrt{5\times2}+\sqrt{3\times 2}}\right):\frac{7}{\sqrt{2\times7}}= [/tex]


[tex]\displaystyle\\ =\left(\frac{\sqrt{3}\times\sqrt{7}+\sqrt{2}\times\sqrt{7}}{\sqrt{3}\times\sqrt{2}+ \sqrt{2}\times\sqrt{2}}+\frac{\sqrt{5}\times\sqrt{7}+\sqrt{3}\times\sqrt{7}}{\sqrt{5}\times\sqrt{2}+\sqrt{3}\times\sqrt{2}}\right):\frac{7}{\sqrt{2\times7}}=\\\\\\ =\left(\frac{\sqrt{7}(\sqrt{3}+\sqrt{2})}{\sqrt{2}(\sqrt{3}+ \sqrt{2})}+ \frac{\sqrt{7}(\sqrt{5}+\sqrt{3})}{\sqrt{2}(\sqrt{5}+\sqrt{3})}\right):\frac{7}{\sqrt{2}\times \sqrt{7}}= [/tex]


[tex]\displaystyle\\ =\left(\frac{\sqrt{7}}{\sqrt{2}}+ \frac{\sqrt{7}}{\sqrt{2}}\right)\times \frac{\sqrt{2}\times \sqrt{7}}{7}=\\\\\\ =\frac{2\sqrt{7}}{\sqrt{2}}\times \frac{\sqrt{2}\times \sqrt{7}}{7}=\\\\\\ =\frac{2\sqrt{7}\times \sqrt{2}\times \sqrt{7}}{7\sqrt{2}}=\frac{2\sqrt{7}\times \sqrt{7}}{7}= \frac{2\times 7}{7}=\boxed{\bf 2} [/tex]