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Question 1

Question 2

Решите задачу:

$5\sqrt{21}+\sqrt{\frac{3}{7}}-9\sqrt{\frac{7}{3}}=\frac{5\sqrt{21}*\sqrt{21}+\sqrt{3}*\sqrt{3}-9\sqrt{7}*\sqrt{7}}{\sqrt{21}}=\frac{5*21+3-9*7}{\sqrt{21}}=\frac{45}{\sqrt{21}}=\frac{15\sqrt{21}}{7}$

$(\sqrt{2}+\sqrt{3a}+\sqrt{3a+2})(\sqrt{2}+\sqrt{3a}-\sqrt{3a+2})=(\sqrt{2}+\sqrt{3a})^2-(\sqrt{3a+2})^2=2+2\sqrt{6a}+3a-3a-2=2\sqrt{6a}$

$\frac{8+2\sqrt{15}}{\sqrt{3}+\sqrt{5}}=\frac{(8+2\sqrt{15})(\sqrt{3}-\sqrt{5})}{(\sqrt{3}+\sqrt{5})(\sqrt{3}-\sqrt{5})}=\frac{(2\sqrt{45}-2\sqrt{75}+8\sqrt{3}-8\sqrt{5})}{3-5}=\frac{(6\sqrt{5}-10\sqrt{3}+8\sqrt{3}-8\sqrt{5})}{3-5}=\frac{-2\sqrt{5}-2\sqrt{3}}{-2}=\sqrt{5}+\sqrt{3}$

$\frac{7a-9b}{\sqrt{7a}+3\sqrt{b}}=\frac{(\sqrt{7a}+3\sqrt{b})(\sqrt{7a}-3\sqrt{b})}{\sqrt{7a}+3\sqrt{b}}=\sqrt{7a}-3\sqrt{b}$

$\frac{\sqrt{a^4b^2+2a^3b^3+a^2b^4}}{\sqrt{(a-b)(a^2-b^2)}}-\frac{\sqrt{a^3b^2+a^2b^3-1}}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}=\frac{\sqrt{a^2b^2(a^2+2ab+b^2)}}{\sqrt{(a-b)(a-b)(a+b)}}-\frac{\sqrt{a^2b^2(a+b)-1}}{(\sqrt{a})^2-(\sqrt{b})^2}=\frac{ab(a+b)}{(a-b)\sqrt{a+b}}-\frac{\sqrt{(ab\sqrt{a+b}-1)(ab\sqrt{a+b}+1)}}{a-b}}=\frac{ab\sqrt{a+b}-\sqrt{(ab\sqrt{a+b}-1)(ab\sqrt{a+b}+1)}}{a-b}=\frac{ab\sqrt{a+b}-\sqrt{a^2b^2(a+b)-1}}{a-b}$

$\sqrt{9+4\sqrt{2}}-\sqrt{9-4\sqrt{2}}=\sqrt{1^2+2*1*2\sqrt{2}+(2\sqrt{2})^2}-\sqrt{1^2-2*1*2\sqrt{2}+(2\sqrt{2})^2}=\sqrt{(1+2\sqrt{2})^2}-\sqrt{(1-2\sqrt{2})^2}=1+2\sqrt{2}+1-2\sqrt{2}=2$