(a^2-ax+bx-ab)/(a^2+ax-bx-ab)*(a^2+ax+ab+bx)/(a^2-ax-ab+bx)

1 answer:

8 0

$\frac{a^2-ax+bx-ab}{a^2+ax-bx-ab} \cdot \frac{a^2+ax+ab+bx}{a^2-ax-ab+bx} = \frac{1}{a(a+x)-b(x+a)} \cdot (a(a+x)+b(a+x)) = \frac{1}{a-b} \cdot (a+b) = \frac{a+b}{a-b}.$

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