Question

Cho a+b=1. Tính giá trị biểu thức C = 2(a³+b³) - 3(a²+b²)

Answer 1

\(C=2\left(a^3+b^3\right)-3\left(a^2+b^2\right)\)

\(=2\left(a+b\right)\left(a^2-ab+b^2\right)-3\left[\left(a+b\right)^2-2ab\right]\)

\(=2\left(a+b\right)\left(a^2-ab+b^2\right)-3\left(a+b\right)^2+6ab\)

\(=2\left(a^2-ab+b^2\right)-3+6ab\) (vì \(a+b=1\))

\(=2a^2-2ab+2b^2-3+6ab\)

\(=2a^2+4ab+2b^2-3\)

\(=2\left(a^2+2ab+b^2\right)-3\)

\(=2\left(a+b\right)^2-3\)

\(=2-3\) (vì \(a+b=1\))

\(=-1\)