# Can you show this? (sqrt(2x^2+1))/(x)=sqrt(2+1/x^2).This seems like basic algebra, but I can't really manage to deal with the sqrt sign and prove this.

Can you show this?
Show that
$\frac{\sqrt{2{x}^{2}+1}}{x}=\sqrt{2+1/{x}^{2}}.$
This seems like basic algebra, but I can't really manage to deal with the sqrt sign and prove this.
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Andrejkoxg
I think it's wrong.
Try $x=-1$.
By the way,
$\frac{\sqrt{2{x}^{2}+1}}{x}=\sqrt{2+\frac{1}{{x}^{2}}}$
for x>0 and
$\frac{\sqrt{2{x}^{2}+1}}{x}=-\sqrt{2+\frac{1}{{x}^{2}}}$
for x<0.
The both equalities we can write so:
$\sqrt{2+\frac{1}{{x}^{2}}}=\frac{\sqrt{2{x}^{2}+1}}{\sqrt{{x}^{2}}}=\frac{\sqrt{2{x}^{2}+1}}{|x|}$

Frida Faulkner
1) $x>0,$ the equality is OK.
2) $x<0$, LHS is negative, RHS positive.
LHS: $\frac{\sqrt{2{x}^{2}+1}}{x}=$
$\frac{\sqrt{{x}^{2}\left(2+\frac{1}{{x}^{2}}\right)}}{x}=$
$\frac{|x|\sqrt{2+\frac{1}{{x}^{2}}}}{x}=$
$\frac{|x|}{x}\sqrt{2+\frac{1}{{x}^{2}}}.$
Note:
$\frac{|x|}{x}=1$ for $x>0;$
$\frac{|x|}{x}=-1$ for $x<0.$