Question 383283
Use the common denominator, {{{(x+y)(x-y)=x^2-y^2}}}
{{{(2x-y)/(x-y)=((2x-y)(x+y))/((x-y)(x+y))}}}
{{{(2x-y)/(x-y)=(2x^2+2xy-xy-y^2)/((x-y)(x+y))}}}
{{{(2x-y)/(x-y)=(2x^2+xy-y^2)/ ((x-y)(x+y))}}}
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{{{(-x-2y)/(x+y)=((-x-2y)(x-y))/((x-y)(x+y))}}}
{{{(-x-2y)/(x+y)= (-x^2+xy-2xy+2y^2)/((x-y)(x+y))}}}
{{{(-x-2y)/(x+y)= (-x^2-xy+2y^2)/((x-y)(x+y))}}}
So then,
{{{(2x-y)/(x-y)+(-x-2y)/(x+y)=(2x^2+xy-y^2)/ ((x-y)(x+y))+(-x^2-xy+2y^2)/((x-y)(x+y))}}}
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{{{(2x-y)/(x-y)+(-x-2y)/(x+y)=(2x^2+xy-y^2-x^2-xy+2y^2)/((x-y)(x+y))}}}
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{{{(2x-y)/(x-y)+(-x-2y)/(x+y) =(x^2+y^2)/((x-y)(x+y))}}}
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{{{(2x-y)/(x-y)+(-x-2y)/(x+y) =highlight((x^2+y^2)/(x^2-y^2))}}}