Question

Simplifique a fração (a²+b²-c²)²-(a²-b²+c²)²
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4ab²+4abc

Answer 1

Resposta:

$\frac{a(b-c)}{b}$

Passo-a-passo:

[ ] é a forma de destacar o par, para a equação do quadrado perfeito

$\frac{(a^2 + [b^2 - c^2])^2 - ([a^2 - b^2] - c^2)^2}{4ab^2-4abc} \\ =\frac{(a^2)^2+2a^2(b^2-c^2)+(b^2-c^2)^2-((a^2-b^2)+2c^2(a^2-b^2)+(c^2)^2)}{4ab(b+c)} =\\\frac{a^4+2a^2b^2-2a^2c^2+(b^2-c^2)^2-(a^2-b^2)^2-2a^2c^2+2c^2b^2+c^4}{4ab(b+c)} =\\\frac{a^4+2a^2b^2+(b^2)^2-2b^2c^2+(c^2)^2-((a^2)^2-2a^2b^2-(b^2)^2)-2a^2c^2+2c^2b^2-c^4}{4ab(b+c)} =\\\frac{a^4+2a^2b^2-2a^2c^2+b^4-2b^2c^2+c^4-a^4+2a^2b^2-b^4-2a^2b^2+2b^2c^2-c^2}{4ab(b+c)} \\Cortando= \frac{4a^2b^2-4a^2c^2}{4ab(b+c)} =\frac{4a^2(b^2-c^2)}{4ab(b+c)}$$\frac{4a^2((b+c)(b-c)}{4ab(b+c)}=\frac{4a^2(b-c)}{4ab}= \frac{a(b-c)}{b}$