Question

simplifique a²-b²+c²+2ac/a²+b²+c²+2ab+2bc+2ac

Answer 1

$\displaystyle \sf \frac{a^2+2ac+c^2-b^2}{a^2+b^2+c^2+2ab+2bc+2ac} \\\\\\ \frac{(a+c)^2-b^2 }{(a+b+c)^2} \\\\\\\ \text{Lembrete :} \\\\ x^2-y^2 = (x+y)(x-y)\\\\ observe\ que \ no \ numerador \ temos : \\\\ (a+c)^2-b^2 = ([a+c]+b)([a+c]-b) \\\\ Da{\'i}}: \\\\\ \frac{(a+c)^2-b^2 }{(a+b+c)^2} \\\\\\ \frac{(a+c+b)(a+c-b)}{(a+b+c)^2 } \to \frac{a+c-b}{a+c+b} \\\\\\ Portanto : \\\\ \boxed{\sf \ \frac{a^2+2ac+c^2-b^2}{a^2+b^2+c^2+2ab+2bc+2ac} = \frac{a+c-b}{a+c+b} \ }\checkmark$