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disa [49]
3 years ago
11

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

Mathematics
1 answer:
Vanyuwa [196]3 years ago
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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