Question

a^2+2bc-b^2-c^2/b^2-a^2-c^2+2ac

Answer 1

$a^2+2bc-b^2-c^2 = a^2-(b^2-2bc+c^2) = a^2-(b-c)^2 = \ =(a-(b-c))(a+(b-c))= (a-b+c)(a+b-c)$

$b^2-a^2-c^2+2ac = b^2-(a^2-2ac+c^2) = b^2-(a-c)^2 = \ = (b-(a-c))(b+(a-c))= (b-a+c)(b+a-c) = \ = (-a+b+c)(a+b-c) = -(a-b-c)(a+b-c)$

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$frac{a^2+2bc-b^2-c^2}{b^2-a^2-c^2+2ac} = frac{(a-b+c)(a+b-c)}{-(a-b-c)(a+b-c)} = -frac{a-b+c}{a-b-c}$

Answer 2

огромное спасибо!!!!!!!!