Question 39381
{{{ (m+n)^(-1)/(m^(-2)-n^(-2)) }}}


lets look at the denominator first:
{{{ m^(-2)-n^(-2) }}}
{{{ 1/m^2-1/n^2 }}}
{{{ (n^2/(n^2m^2))-(m^2/(m^2n^2)) }}}
{{{ ((n^2-m^2)/(n^2m^2)) }}}


so we end up with:
{{{ (m+n)^(-1)/((n^2-m^2)/(n^2m^2)) }}}
{{{ (m+n)^(-1)*((n^2m^2)/(n^2-m^2)) }}}
{{{ (1/(m+n))*((n^2m^2)/(n^2-m^2)) }}}
{{{ (1/(m+n))*((n^2m^2)/((n-m)(n+m))) }}}
{{{ (1/(n+m))*((n^2m^2)/((n-m)(n+m))) }}}
{{{ (n^2m^2)/((n-m)(n+m)(n+m)) }}}
{{{ (n^2m^2)/((n-m)(n+m)^2) }}}


jon.