Question 18865
{{{(x^-2-y^-2)/(x^-1-y^-1)}}}
A negative power means that you can put the reciprocal of that number
ie. x^-2 is the same as (1/x^2)
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keeping that in mind,this expression becomes
{{{((1/x^2)-(1/y^2))/((1/x)-(1/y))}}}
Taking LCM this becomes,
{{{((y^2-x^2)/(x^2*y^2))/((y-x)/(xy))}}}
Now we get,
{{{((y^2-x^2)/(x^2*y^2))*((xy)/(y-x))}}}
**(y^2-x^2)=(y-x)(y+x).Substituting this we get,
{{{ ((y-x)(y+x)/(x^2*y^2))*((xy)/(y-x)) }}}
**(x^2*y^2)=(xy)(xy).Substituting this we get,
{{{ ((y-x)(y+x)/(xy*xy))*((xy)/(y-x)) }}}
Now cancelling out terms we get,
{{{ ((y+x)/(xy)) }}}
And this is the final answer.
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Hope this helps,
Prabhat