Question

cho a+b=1. Tính gia trị biểu thức:

\(\text{a}=\text{a}^3+b^3+3\text{a}b\)

\(B=4\left(\text{a}^3+b^3\right)-6\left(\text{a}^2+b^2\right)\)

Answer 1

\(A=a^3+b^3+3ab\)

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

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

\(=a^2+2ab+b^2\)

\(=\left(a+b\right)^2=1^2=1\)

\(B=4\left(a^3+b^3\right)-6\left(a^2+b^2\right)\)

\(=4\left(a+b\right)\left(a^2-ab+b^2\right)-6a^2-6b^2\)

\(=4\left(a^2-ab+b^2\right)-6a^2-6b^2\)

\(=4a^2-4ab+4b^2-6a^2-6b^2\)

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

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

\(=-2\left(a+b\right)^2=-2.1^2=-2\)