Question

\(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}\)

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Answer 1

\(\frac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}\)

\(=\frac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}\)

\(=\frac{\left(a+b-c\right)\left(a+b+c\right)}{\left(a+c-b\right)\left(a+b+c\right)}\)

\(=\frac{a+b-c}{a+c-b}\left(a+b+c\ne0\right)\)