Question

Cho a + b +c = 0 , a² + b² + c² = 1

Tính a⁴ + b² + c⁴

Answer 1

\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\Leftrightarrow ab+bc+ac=-\frac{1}{2}\)

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

Ta có ; \(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=a^4+b^4+c^4+\frac{1}{2}=1\)

\(\Rightarrow a^4+b^4+c^4=\frac{1}{2}\)