$$ \begin{array}{ll} ext { a) } \sqrt[3]{7} \cdot \sqrt[5]{7} & ext { d) }\left(x^{\frac{5}{4}}: y^{-\frac{5}{8}} ight)^{-\frac{4}{5}} \ ext { b) } x^{-\frac{1}{2}} \cdot x^{4} & ext { e) } \sqrt[3]{\sqrt{2}} \ ext { c) }(2 a)^{\frac{5}{4}}:(2 a)^{\frac{3}{4}} & ext { f) } \frac{x}{\sqrt[3]{x^{2}} \cdot \sqrt[4]{x}}\end{array} $$

Question

$$ \begin{array}{ll}	ext { a) } \sqrt[3]{7} \cdot \sqrt[5]{7} & 	ext { d) }\left(x^{\frac{5}{4}}: y^{-\frac{5}{8}}ight)^{-\frac{4}{5}} \ 	ext { b) } x^{-\frac{1}{2}} \cdot x^{4} & 	ext { e) } \sqrt[3]{\sqrt{2}} \ 	ext { c) }(2 a)^{\frac{5}{4}}:(2 a)^{\frac{3}{4}} & 	ext { f) } \frac{x}{\sqrt[3]{x^{2}} \cdot \sqrt[4]{x}}\end{array} $$

UpStudy AI · Accepted Answer

Let's simplify each expression one by one.

### a) $$ \sqrt[3]{7} \cdot \sqrt[5]{7} $$

We can express the roots as exponents:
$$
\sqrt[3]{7} = 7^{\frac{1}{3}}, \quad \sqrt[5]{7} = 7^{\frac{1}{5}}
$$
Now, we can add the exponents:
$$
7^{\frac{1}{3}} \cdot 7^{\frac{1}{5}} = 7^{\frac{1}{3} + \frac{1}{5}}
$$
To add the fractions, we find a common denominator, which is 15:
$$
\frac{1}{3} = \frac{5}{15}, \quad \frac{1}{5} = \frac{3}{15}
$$
Thus,
$$
\frac{1}{3} + \frac{1}{5} = \frac{5}{15} + \frac{3}{15} = \frac{8}{15}
$$
So,
$$
\sqrt[3]{7} \cdot \sqrt[5]{7} = 7^{\frac{8}{15}}
$$

### b) $$ x^{-\frac{1}{2}} \cdot x^{4} $$

Using the property of exponents $$ a^m \cdot a^n = a^{m+n} $$:
$$
x^{-\frac{1}{2}} \cdot x^{4} = x^{-\frac{1}{2} + 4} = x^{-\frac{1}{2} + \frac{8}{2}} = x^{\frac{7}{2}}
$$

### c) $$ (2 a)^{\frac{5}{4}}:(2 a)^{\frac{3}{4}} $$

This is a division of exponents:
$$
(2 a)^{\frac{5}{4}} : (2 a)^{\frac{3}{4}} = (2 a)^{\frac{5}{4} - \frac{3}{4}} = (2 a)^{\frac{2}{4}} = (2 a)^{\frac{1}{2}} = \sqrt{2 a}
$$

### d) $$ \left(x^{\frac{5}{4}}: y^{-\frac{5}{8}}\right)^{-\frac{4}{5}} $$

First, simplify the division:
$$
x^{\frac{5}{4}} : y^{-\frac{5}{8}} = x^{\frac{5}{4}} \cdot y^{\frac{5}{8}} = \frac{x^{\frac{5}{4}}}{y^{-\frac{5}{8}}} = x^{\frac{5}{4}} \cdot y^{\frac{5}{8}}
$$
Now raise this to the power of $$-\frac{4}{5}$$:
$$
\left(x^{\frac{5}{4}} \cdot y^{\frac{5}{8}}\right)^{-\frac{4}{5}} = x^{-\frac{4}{5} \cdot \frac{5}{4}} \cdot y^{-\frac{4}{5} \cdot \frac{5}{8}} = x^{-1} \cdot y^{-\frac{1}{2}} = \frac{1}{x \sqrt{y}}
$$

### e) $$ \sqrt[3]{\sqrt{2}} $$

Express the square root as an exponent:
$$
\sqrt{2} = 2^{\frac{1}{2}} \quad \Rightarrow \quad \sqrt[3]{\sqrt{2}} = \sqrt[3]{2^{\frac{1}{2}}} = 2^{\frac{1}{2} \cdot \frac{1}{3}} = 2^{\frac{1}{6}}
$$

### f) $$ \frac{x}{\sqrt[3]{x^{2}} \cdot \sqrt[4]{x}} $$

Express the roots as exponents:
$$
\sqrt[3]{x^{2}} = x^{\frac{2}{3}}, \quad \sqrt[4]{x} = x^{\frac{1}{4}}
$$
Now multiply the denominators:
$$
\sqrt[3]{x^{2}} \cdot \sqrt[4]{x} = x^{\frac{2}{3}} \cdot x^{\frac{1}{4}} = x^{\frac{2}{3} + \frac{1}{4}}
$$
Finding a common denominator (12):
$$
\frac{2}{3} = \frac{8}{12}, \quad \frac{1}{4} = \frac{3}{12} \quad \Rightarrow \quad \frac{2}{3} + \frac{1}{4} = \frac{8}{12} + \frac{3}{12} = \frac{11}{12}
$$
Thus,
$$
\sqrt[3]{x^{2}} \cdot \sqrt[4]{x} = x^{\frac{11}{12}}
$$
Now, we can simplify:
$$
\frac{x}{\sqrt[3]{x^{2}} \cdot \sqrt[4]{x}} = \frac{x}{x^{\frac{11}{12}}} = x^{1 - \frac{11}{12}} = x^{\frac{1}{12}}
$$

### Summary of Results
- a) $$ 7^{\frac{8}{15}} $$
- b) $$ x^{\frac{7}{2}} $$
- c) $$ \sqrt{2 a} $$
- d) $$ \frac{1}{x \sqrt{y}} $$
- e) $$ 2^{\frac{1}{6}} $$
- f) $$ x^{\frac{1}{12}} $$
a) $$ 7^{\frac{8}{15}} $$ b) $$ x^{\frac{7}{2}} $$ c) $$ \sqrt{2 a} $$ d) $$ \frac{1}{x \sqrt{y}} $$ e) $$ 2^{\frac{1}{6}} $$ f) $$ x^{\frac{1}{12}} $$

Solución de tutoría real

Responder

Solución

Extra Insights

preguntas relacionadas

Latest Algebra Questions