Задача 53498 Подскажите, как решить....
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[m] (\sqrt{a}+\sqrt{b})^2-4b=(\sqrt{a}+ \sqrt{b})^2-(2\sqrt{b})^2=[/m]
[m]=(\sqrt{a}+ \sqrt{b}-2\sqrt{b})\cdot (\sqrt{a}+ \sqrt{b}+2\sqrt{b})=(\sqrt{a}- \sqrt{b})\cdot (\sqrt{a}+ 3\sqrt{b})[/m]
[m]\frac{1}{\sqrt{b}}+\frac{3}{\sqrt{a}}=\frac{\sqrt{a}+3\sqrt{b}}{\sqrt{a}\cdot \sqrt{b}}[/m]
[m] \frac{(\sqrt{a}+\sqrt{b})^2-4b}{(a-b):(\frac{1}{\sqrt{b}}+\frac{3}{\sqrt{a}})}= \frac{(\sqrt{a}-\sqrt{b})\cdot (\sqrt{a}+3\sqrt{b})}{(a-b):(\frac{\sqrt{a}+3\sqrt{b}}{\sqrt{a}\cdot \sqrt{b}})}= \frac{(\sqrt{a}-\sqrt{b})\cdot (\sqrt{a}+3\sqrt{b})}{(\sqrt{a}-\sqrt{b})\cdot (\sqrt{a}+\sqrt{b})\cdot \frac{\sqrt{a}\cdot \sqrt{b}}{\sqrt{a}+3\sqrt{b}}}= \frac{(\sqrt{a}+3\sqrt{b})^2}{ (\sqrt{a}+\sqrt{b})\cdot \sqrt{a}\cdot \sqrt{b}}[/m]
[m]a+9b+6\sqrt{ab}=(\sqrt{a}+3\sqrt{b})^2[/m]
[m]\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{a}}=\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}\cdot \sqrt{b}}[/m]
[m]\frac{a+9b+6\sqrt{ab}}{\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{a}}}=\frac{(\sqrt{a}+3\sqrt{b})^2}{\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}\cdot \sqrt{b}}}[/m]
[m] \frac{(\sqrt{a}+3\sqrt{b})^2}{ (\sqrt{a}+\sqrt{b})\cdot \sqrt{a}\cdot \sqrt{b}}:\frac{(\sqrt{a}+3\sqrt{b})^2}{\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}\cdot \sqrt{b}}}= \frac{(\sqrt{a}+3\sqrt{b})^2}{ (\sqrt{a}+\sqrt{b})\cdot \sqrt{a}\cdot \sqrt{b}}\cdot \frac{\sqrt{a}\cdot \sqrt{b}}{(\sqrt{a}+3\sqrt{b})^2}=\frac{1}{\sqrt{a}+\sqrt{b}}[/m]