Question

Cho a+b+c=0 và a²+b²+c²=1. Tính M=a⁴+b⁴+c⁴

Answer 1

Từ \(a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=0\)

Hay \(a^2+b^2+c^2+2ab+2bc+2ac=0\)

\(\Leftrightarrow1+2ab+2bc+2ac=0\)

\(\Leftrightarrow2ab+2bc+2ac=-1\Leftrightarrow ab+bc+ac=-\dfrac{1}{2}\)

\(\Leftrightarrow\left(ab+bc+ac\right)^2=\dfrac{1}{4}\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=\dfrac{1}{4}\)

Thay \(a+b+c=0\) ta có: \(a^2b^2+b^2c^2+a^2c^2=\dfrac{1}{4}\)

Trở lại bài toán

\(M=a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+b^2c^2+a^2c^2\right)=1-2.\dfrac{1}{4}=1-\dfrac{1}{2}=\dfrac{1}{2}\)