(x^2-1)^1/2 - (x^2-1)^-1/2(2-x^2) all divided by x^2-1

\[\frac{(x^2-1)^{1/2}- (x^2-1)^{-1/2}(2-x^2)}{x^2-1}\] like that?

yes and i have solve for x

and it =0 of course

you might multiply numerator and denominator by \[(x^2-1)^{1/2} \] to clear the fraction but it is not an equation so you cannot solve for x

oh = 0. ok. still clear the fraction get \[\frac{(x^2-1)^{1/2}- (x^2-1)^{-1/2}(2-x^2)}{x^2-1}\times \frac{(x^2-1)^{1/2}}{(x^2-1)^{1/2}}\] \[=\frac{x^2-1-(2-x^2)}{(x^2-1)^{3/2}}\] \[=\frac{2x^2-3}{(x^2-1)^{3/2}}\]

if this equals zero it means the numerator is equal zero so you can solve via \[2x^2-3=0\] \[x^2=\frac{3}{2}\] \[x=\pm\sqrt{\frac{3}{2}}\]

I know that \[(x ^{2}-1)^{1\div2} \] cancels out but where does the -1/2 go because wouldn't that be -1/4 exponent

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