The given rational fraction is
[tex]\frac{\sqrt{49x^8}}{\sqrt{7x^3}}[/tex]
We will simplify it
First, simplify the numerator
Find the square root of 49 and the square root of x^8
[tex]\begin{gathered} \sqrt{49}=7 \\ \sqrt{x^8}=x^{\frac{8}{2}}=x^4 \\ \sqrt{49x^8}=7x^4 \end{gathered}[/tex]
Now, simplify the denominator
Find the square root of x^3
[tex]\begin{gathered} x^3=x\times x\times x \\ \sqrt{x^3}=\sqrt{x^2\times x}=x\sqrt{x} \\ \sqrt{7x^3}=x\sqrt{7x} \end{gathered}[/tex]
The fraction is
[tex]\frac{7x^4}{x\sqrt{7x}}[/tex]
Simplify x^4 up with x down by subtracting their powers
[tex]\frac{7x^{4-1}}{\sqrt{7x}}=\frac{7x^3}{\sqrt{7x}}[/tex]
Multiply up and down by root 7x to rationalize the denominator
[tex]\begin{gathered} \frac{7x^3}{\sqrt{7x}}\times\frac{\sqrt{7x}}{\sqrt{7x}}= \\ \\ \frac{7x^3(\sqrt{7x})}{7x}= \end{gathered}[/tex]
Simplify 7x^3 up with 7x down
[tex]\frac{7x^3}{7x}=\frac{7}{7}x^{3-1}=1(x^2)=x^2[/tex]
Then the simplest form is
[tex]x^2\sqrt{7x}[/tex]
The answer is
[tex]x^2\sqrt{7x}[/tex]
