Question

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

Answer 1

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

Answer 2

yes and i have solve for x

Answer 3

and it =0 of course

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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