Question

Cho a+b+c=5

chứng minh rằng:

a³+b³+c³-3abc/a²+b²+c²-ab-bc-ac

Answer 1

Ta có: \(\frac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-bc-ac}\)

\(=\frac{\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc}{a^2+b^2+c^2-ab-bc-ac}\)

\(=\frac{\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)}{a^2+b^2+c^2-ab-bc-ac}\)

\(=\frac{\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)}{a^2+b^2+c^2-ab-bc-ac}\)

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

\(=a+b+c\)

=5