# Why does $\frac{1}{x\sqrt{x}}= \frac {\sqrt{x}}{x^2}$?

A homework question recently asked for me to simplify:

$\frac{1}{\sqrt{7}} \div {7}$

It's easy to see that this becomes

$\frac{1}{7\sqrt{7}}$

But according to wolfram alpha this is also equal to $\frac{\sqrt{7}}{49}$.

What sequence of steps can I use to get the second representation of this quantity from the first?

• $$\frac{1}{7\sqrt{7}}\cdot \frac{\sqrt{7}}{\sqrt{7}}.$$ – Daniel Fischer Dec 8 '13 at 14:52
• Multiply both numerator and denominator bt Sqrt[7] and simplify the denominator [7 Sqrt(7) Sqrt(7)] = 7 7 = 49 – Claude Leibovici Dec 8 '13 at 14:52
• If you replace 7 by x, you have the same solution – Claude Leibovici Dec 8 '13 at 15:04

$\dfrac{1}{\sqrt{7}}\div7=\dfrac{1}{7\sqrt{7}}=\dfrac{1}{7\sqrt{7}}\cdot\dfrac{\sqrt{7}}{\sqrt{7}}=\dots$
$$\frac{1}{7\sqrt{7}} =\frac{\sqrt{7}}{7\sqrt{7}\cdot\sqrt{7}}=\frac{\sqrt{7}}{7\cdot7}=\frac{\sqrt{7}}{49}$$