Question

Rationalize the denominator of
\[\frac{sqrt{x^2}}{sqrt{x}+1}\]

Answer 1

\[\frac{\sqrt{x^2}}{\sqrt{x}+1}\]

Answer 2

I got \[\frac{x^3-x}{x-1}\]

Answer 3

\[\frac{\sqrt{x^2}}{\sqrt{x}+1} = \frac{x}{\sqrt{x}+1}=\frac{x(\sqrt{x}-1)}{(\sqrt{x}+1)(\sqrt{x}-1)}=\frac{x\sqrt{x}-x}{x-1}\]

Answer 4

I see what I did wrong thanks @RolyPoly

Answer 5

Note that \[\sqrt{x^2} = x\]

Answer 6

Okay, I noted that :). Thank You once again!

Answer 7

welcome