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Show that $$\frac{\sqrt{2x^2+1}}{x}=\sqrt{2+1/x^2}.$$

This seems like basic algebra, but I can't really manage to deal with the sqrt sign and prove this.

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closed as off-topic by Aqua, mfl, Simply Beautiful Art, Did, Math Lover Oct 17 '17 at 17:46

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  • $\begingroup$ Can you fix the formatting please. $\endgroup$ – Joe Oct 17 '17 at 15:10
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    $\begingroup$ If you assume that $x$ is a positive real number, then $\frac{\sqrt{2x^2+1}}{x}=\frac{\sqrt{2x^2+1}}{\sqrt{x^2}}=\sqrt{\frac{2x^2+1}{x^2}}=\sqrt{2\frac{x^2}{x^2}+\frac{1}{x^2}}$ This however does not work going from the first to the second expression if $x$ is not positive. Going from the second to the third expression should cause concern as well, but is valid again since just positive numbers. $\endgroup$ – JMoravitz Oct 17 '17 at 15:12
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I think it's wrong.

Try $x=-1$.

By the way, $$\frac{\sqrt{2x^2+1}}{x}=\sqrt{2+\frac{1}{x^2}}$$ for $x>0$ and

$$\frac{\sqrt{2x^2+1}}{x}=-\sqrt{2+\frac{1}{x^2}}$$ for $x<0$.

The both equalities we can write so: $$\sqrt{2+\frac{1}{x^2}}=\frac{\sqrt{2x^2+1}}{\sqrt{x^2}}=\frac{\sqrt{2x^2+1}}{|x|}$$

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1) $x>0,$ the equality is OK.

2) $x < 0$, LHS is negative, RHS positive.

LHS: $\dfrac{\sqrt{2x^2 +1} }{x}=$

$\dfrac {\sqrt{x^2(2+\dfrac{1}{x^2})}}{x} =$

$\dfrac{|x|\sqrt{2+\dfrac{1}{x^2}}}{x} =$

$\dfrac{|x|}{x}\sqrt{2+\dfrac{1}{x^2}}.$

Note:

$\dfrac{|x|}{x} = 1$ for $x>0;$

$\dfrac{|x|}{x} = -1$ for $x<0.$

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