Question

A1, subpunctul e).
Mulțumesc anticipat.

Answer 1

$\left(\sqrt[4]{a}+\sqrt[4]{b}\right):\left[\left(\sqrt[10]{\dfrac{\sqrt{a^3b^3}}{\sqrt[3]{a^2}\sqrt{b^3}}}\right)^3+\dfrac{\sqrt[8]{b^2a^{20}}}{\sqrt{a^5}}\right] =$

$=\left(\sqrt[4]{a}+\sqrt[4]{b}\right):\left(\dfrac{(ab)^{3\cdot \frac{1}{2}\cdot \frac{1}{10}\cdot 3}}{a^{2\cdot \frac{1}{3}\cdot \frac{1}{10}\cdot 3}\cdot b^{3\cdot \frac{1}{2}\cdot \frac{1}{10}\cdot 3}}+\dfrac{b^{\frac{2}{8}}\cdot a^{\frac{20}{8}}}{a^{\frac{5}{2}}}\right)=\\ \\=\left(\sqrt[4]{a}+\sqrt[4]{b}\right):\left(\dfrac{\left(a\cdot b\right)^{\frac{9}{20}}}{a^{\frac{1}{5}}\cdot b^{\frac{9}{20}}}+\dfrac{b^{\frac{1}{4}}\cdot a^{\frac{5}{2}}}{a^{\frac{5}{2}}}\right)=$

$=\left(\sqrt[4]{a}+\sqrt[4]{b}\right):\left(\dfrac{a^{\frac{9}{20}}\cdot b^{\frac{9}{20}}}{a^{\frac{1}{5}}\cdot b^{\frac{9}{20}}}+b^{\frac{1}{4}}\right)= \\ \\ =\left(\sqrt[4]{a}+\sqrt[4]{b}\right):\left(a^{\frac{9}{20} -\frac{1}{5}}+b^{\frac{1}{4}}\right)=\\ \\ = \left(\sqrt[4]{a}+\sqrt[4]{b}\right):\left(a^{\frac{9-4}{20}}+b^{\frac{1}{4}}\right) =$

$=\left(\sqrt[4]{a}+\sqrt[4]{b}\right):\left(a^{\frac{1}{4}}+b^{\frac{1}{4}}\right)=\\ \\ = \left(\sqrt[4]{a}+\sqrt[4]{b}\right):\left(\sqrt[4]{a}+\sqrt[4]{b}\right) =\\ \\ = \boxed{1}$

Answer 2

Răspuns:

1

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