Question
Perform the operations and simplify
(\sqrt(6x^(3)))/(\sqrt(3x))-:\sqrt(8x^(2))
(\sqrt(6x^(3)))/(\sqrt(3x))-:\sqrt(8x^(2))
Answers
GPT-4o mini
To perform the operations and simplify the expression \(\frac{\sqrt{6x^3}}{\sqrt{3x}} \div \sqrt{8x^2}\), we can start by rewriting the division as multiplication by the reciprocal:
\[
\frac{\sqrt{6x^3}}{\sqrt{3x}} \div \sqrt{8x^2} = \frac{\sqrt{6x^3}}{\sqrt{3x}} \cdot \frac{1}{\sqrt{8x^2}} = \frac{\sqrt{6x^3}}{\sqrt{3x} \cdot \sqrt{8x^2}}.
\]
Next, we can simplify the denominator:
\[
\sqrt{3x} \cdot \sqrt{8x^2} = \sqrt{3 \cdot 8 \cdot x \cdot x^2} = \sqrt{24x^3}.
\]
Now we rewrite the expression:
\[
\frac{\sqrt{6x^3}}{\sqrt{24x^3}}.
\]
Using the property of square roots \(\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}\):
\[
\frac{\sqrt{6x^3}}{\sqrt{24x^3}} = \sqrt{\frac{6x^3}{24x^3}}.
\]
Simplifying the fraction inside the square root:
\[
\frac{6x^3}{24x^3} = \frac{6}{24} = \frac{1}{4}.
\]
The \(x^3\) terms in the numerator and denominator cancel out:
\[
\sqrt{\frac{1}{4}} = \frac{1}{2}.
\]
Thus, the final simplified result is:
\[
\frac{1}{2}.
\]
\[
\frac{\sqrt{6x^3}}{\sqrt{3x}} \div \sqrt{8x^2} = \frac{\sqrt{6x^3}}{\sqrt{3x}} \cdot \frac{1}{\sqrt{8x^2}} = \frac{\sqrt{6x^3}}{\sqrt{3x} \cdot \sqrt{8x^2}}.
\]
Next, we can simplify the denominator:
\[
\sqrt{3x} \cdot \sqrt{8x^2} = \sqrt{3 \cdot 8 \cdot x \cdot x^2} = \sqrt{24x^3}.
\]
Now we rewrite the expression:
\[
\frac{\sqrt{6x^3}}{\sqrt{24x^3}}.
\]
Using the property of square roots \(\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}\):
\[
\frac{\sqrt{6x^3}}{\sqrt{24x^3}} = \sqrt{\frac{6x^3}{24x^3}}.
\]
Simplifying the fraction inside the square root:
\[
\frac{6x^3}{24x^3} = \frac{6}{24} = \frac{1}{4}.
\]
The \(x^3\) terms in the numerator and denominator cancel out:
\[
\sqrt{\frac{1}{4}} = \frac{1}{2}.
\]
Thus, the final simplified result is:
\[
\frac{1}{2}.
\]