Question

Спростите вираз.Даю 50 балов​

Answer 1

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Answer 2

Ответ:

$\Big(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\Big)\cdot \dfrac{1}{a-b}+1:\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{b}}=\\\\\\=\Big(\dfrac{(\sqrt{a}+\sqrt{b})(a-\sqrt{ab}+b)}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\Big)\cdot \dfrac{1}{a-b}+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}=\\\\\\=\dfrac{(\sqrt{a}+\sqrt{b})(a-\sqrt{ab}+b)-\sqrt{ab}(\sqrt{a}+\sqrt{b})}{\sqrt{a}+\sqrt{b}}\cdot \dfrac{1}{a-b}+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}=$

$=\dfrac{(\sqrt{a}+\sqrt{b})(a-2\sqrt{ab}+b)}{\sqrt{a}+\sqrt{b}}\cdot \dfrac{1}{a-b}+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}=\\\\\\=\dfrac{(\sqrt{a}-\sqrt{b})^2}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}=\dfrac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}+\dfrac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}=\\\\\\=\dfrac{\sqrt{a}-\sqrt{b}+2\sqrt{b}}{\sqrt{a}+\sqrt{b}}=\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}=1$