Write as:
#x^2/(x^2-2^2) = x/(x+2)-2/(2-x)#
Using the principle that #color(white)("s") a^2-b^2=(a+b)(a-b)#
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#color(brown)("The sort of cheating bit - not really!")#
Consider #-2/(2-x)#.#color(white)(c)# We need to change this such that the
#color(white)()#
denominator is #x-2# if we whish to use the context of
#color(white)()#
#color(white)("s") a^2-b^2=(a+b)(a-b)#
Suppose we had #+2/(x-2)# which is the format we wish to have. If we use this is it another form of what was originally written. That is; does it have the same value?
#color(green)([2/(x-2)color(red)(xx1)] ->[2/(x-2)color(red)(xx(-1)/(-1))] = -2/((2-x))#
So #+2/(x-2)# has the same value as #-2/(2-x)#
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Write as:
#x^2/(x^2-2^2) = x/(x+2)+2/(x-2)#
This is the same as:
#x^2/(x^2-2^2)=(x xx2)/((x+2)(x-2))#
#x^2/(x^2-2^2)=(2x)/(x^2-2^2)#
As the denominators are the same we can just consider the numerators.
#x^2=2x#
Divide both sides by x
#x=2#
