Очень срочно, помогите пожалуйста..(

№5

72; \sqrt{80}>\sqrt{72}\\4\sqrt{5}>3\sqrt{8}" alt="4\sqrt{5}=\sqrt{16}\sqrt{5}=\sqrt{16*5}=\sqrt{80}\\3\sqrt{8}=\sqrt{9}\sqrt{8}=\sqrt{9*5}=\sqrt{72}\\80>72; \sqrt{80}>\sqrt{72}\\4\sqrt{5}>3\sqrt{8}" align="absmiddle" class="latex-formula">

$7\sqrt{\frac{2}{7}}=\sqrt{49}\sqrt{\frac{2}{7}}=\sqrt{\frac{49*2}{7}}=\sqrt{14}\\\frac{1}{2}\sqrt{56}=\sqrt{\frac{1}{4}}\sqrt{56}=\sqrt{\frac{56}{4}}=\sqrt{14}\\7\sqrt{\frac{2}{7}}=\frac{1}{2}\sqrt{56}$

№6

$\frac{x-25}{\sqrt{x}-5}=\frac{(\sqrt{x})^2-5^2}{\sqrt{x}-5}=\frac{(\sqrt{x}-5)(\sqrt{x}+5)}{\sqrt{x}-5} =\sqrt{x} +5$

$\frac{6+5\sqrt{6}}{\sqrt{6}}=\frac{\sqrt{6}*\sqrt{6}+5*\sqrt{6}}{\sqrt{6}}=\frac{\sqrt{6}*(\sqrt{6}+5)}{\sqrt{6}}=\sqrt{6}+5$

$\frac{c+14\sqrt{c}+49}{c-49}=\frac{(\sqrt{c})^2+2*7*\sqrt{c}+7^2}{(\sqrt{c})^2-7^2}=\frac{(\sqrt{c}+7)^2}{(\sqrt{c}-7)(\sqrt{c}+7)}=\frac{\sqrt{c}+7}{\sqrt{c}-7}$