The sum of a number and 3 divided by 9

Mathematics

1 answer:

polet [3.4K]3 years ago

6 0

Answer:

$\frac{(x+3)}{9}$

Step-by-step explanation:

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((81x^3 y^(-1/3))^1/4)/(x^2 y)^1/4 simplify

malfutka [58]

$\bf a^{\frac{{ n}}{{ m}}} \implies \sqrt[{ m}]{a^{ n}} \qquad \qquad
\sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}
\\\\\\
\left.\qquad \qquad \right.\textit{negative exponents}\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}
\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\qquad \qquad 
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-------------------------------\\\\$

$\bf \cfrac{\left( 81x^3y^{-\frac{1}{3}} \right)^{\frac{1}{4}}}{(x^2y)^{\frac{1}{4}}}\implies \cfrac{81^{\frac{1}{4}}x^{3\cdot {\frac{1}{4}}}y^{-\frac{1}{3}\cdot {\frac{1}{4}}}}{x^{2\cdot {\frac{1}{4}}}y^{\frac{1}{4}}}\implies
\cfrac{81^{\frac{1}{4}}x^{\frac{3}{4}}y^{-\frac{1}{12}}}{x^{\frac{2}{4}} y^{\frac{1}{4}}}$

$\bf \cfrac{81^{\frac{1}{4}}x^{\frac{3}{4}}x^{-\frac{2}{4}}}{ y^{\frac{1}{4}}y^{\frac{1}{12}}}\implies \cfrac{(3^4)^{\frac{1}{4}}x^{\frac{3}{4}-\frac{2}{4}}}{y^{\frac{1}{4}+\frac{1}{12}}}\implies \cfrac{3^{4\cdot \frac{1}{4}}x^{\frac{1}{4}}}{y^{\frac{4}{12}}}\implies \cfrac{3^{\frac{4}{4}}x^{\frac{1}{4}}}{y^{\frac{1}{3}}}
\\\\\\
\cfrac{3\sqrt[4]{x}}{\sqrt[3]{y}}$

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