Question

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Answer 1

Объяснение:

Вот тебе ответ.........

Answer 2

Ответ:

$\displaystyle \Big(a+b^{\frac{3}{2}}:\sqrt{a}\Big)^{\frac{2}{3}}\cdot \Big(\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}}+\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}\Big)^{-\frac{2}{3}}=$

$\displaystyle=\left(a+\frac{b^{\frac{3}{2}}}{\sqrt{a}}\right )^{\frac{2}{3}}\cdot \left(\frac{(\sqrt{a}-\sqrt{b})^2+a}{\sqrt{a}\cdot (\sqrt{a}-\sqrt{b})}\right)^{-\frac{2}{3}}=\frac{(a^{\frac{3}{2}}+b^{\frac{3}{2}})^{\frac{2}{3}}}{(a^{\frac{1}{2}})^{\frac{2}{3}}}\cdot \left (\frac{\sqrt{a}\cdot (\sqrt{a}-\sqrt{b})}{a-2\sqrt{a}\cdot \sqrt{b}+b+a}\right )^{\frac{2}{3}}=$

$\displaystyle =\frac{(a^{\frac{3}{2}}+b^{\frac{3}{2}})^{\frac{2}{3}}}{(a^{\frac{1}{2}})^{\frac{2}{3}}}\cdot \left (\frac{\sqrt{a}\cdot (\sqrt{a}-\sqrt{b})}{2a-2\sqrt{ab}+b}\right )^{\frac{2}{3}}=\\\\\\=\frac{(a^{\frac{1}{2}}+b^{\frac{1}{2}})^{\frac{2}{3}}(a-a^{\frac{1}{2}}b^{\frac{1}{2}}+b)^{\frac{2}{3}}}{a^{\frac{1}{3}}}\cdot \frac{a^{\frac{1}{3}}\cdot (\sqrt{a}-\sqrt{b})^{\frac{2}{3}}}{(\, 2a-2\sqrt{ab}+b\, )^{\frac{2}{3}}}}}=$

$\displaystyle =\left(\frac{(\sqrt{a}+\sqrt{b})(a-\sqrt{ab}+b)}{1}\cdot \frac{(\sqrt{a}-\sqrt{b})}{2a-2\sqrt{ab}+b}\right )^{\frac{2}{3}}=\left (\frac{(a-b)(a-\sqrt{ab}+b)}{2a-2\sqrt{ab}+b}\right )^{\frac{2}{3}}=$

$\displaystyle =\sqrt[3]{\frac{(\, (a-b)(a-\sqrt{ab}+b)\, )^2}{(\, 2a-2\sqrt{ab}+b\, )^2}}=\sqrt[3]{\frac{(\, a^2-b^2-a\sqrt{ab}+b\sqrt{ab}\, )^2}{(\, 2a-2\sqrt{ab}+b\, )^2}}=\\\\\\=\sqrt[3]{\left (\frac{a^2-b^2-\sqrt{a^3b}+\sqrt{ab^3}}{2a-2\sqrt{ab}+b}\right )^2}$