Question

$( \sqrt{x { {}^{3} } } \times \sqrt[3]{x {}^{5} } ) \div \sqrt[5]{x {}^{3 } } \: \times \sqrt[30]{x {}^{77} }$
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Answer 1

Answer:

$\huge\boxed{\left(\sqrt{x^3}\cdot\sqrt[3]{x^5}\right):\sqrt[5]{x^3}\cdot\sqrt[30]{x^{77}}=\sqrt[15]{x^{77}}}$

Step-by-step explanation:

$a^\frac{m}{n}=\sqrt[n]{a^m}$

$\left(\sqrt{x^3}\cdot\sqrt[3]{x^5}\right):\sqrt[5]{x^3}\cdot\sqrt[30]{x^{77}}$

$\sqrt{x^3}=x^\frac{3}{2}\\\\\sqrt[3]{x^5}=x^\frac{5}{3}\\\\\sqrt[5]{x^3}=x^\frac{3}{5}\\\\\sqrt[30]{x^{77}}=x^\frac{77}{30}$

$\left(\sqrt{x^3}\cdot\sqrt[3]{x^5}\right):\sqrt[5]{x^3}\cdot\sqrt[30]{x^{77}}=\left(x^\frac{3}{2}\cdot x^\frac{5}{3}\right):x^\frac{3}{5}\cdot x^\frac{77}{30}\\\\=x^{\frac{3}{2}+\frac{5}{3}-\frac{3}{5}+\frac{77}{30}}=(*)\\\\\dfrac{3}{2}+\dfrac{5}{3}-\dfrac{3}{5}+\dfrac{77}{30}=\dfrac{3\cdot15}{2\cdot15}+\dfrac{5\cdot10}{3\cdot10}-\dfrac{3\cdot6}{5\cdot6}+\dfrac{77}{30}=\dfrac{45}{30}+\dfrac{50}{30}-\dfrac{18}{30}+\dfrac{77}{30}$

$=\dfrac{45+50-18+77}{30}=\dfrac{154}{30}=\dfrac{154:2}{30:2}=\dfrac{77}{15}=5\dfrac{2}{15}$

$(*)=x^\frac{77}{15}=\sqrt[15]{x^{77}}$