Question

cho a+b+c=0;a^2+b^2+c^2=2. tính a^4+b^4+c^4

Answer 1

\(a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=0\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\Leftrightarrow1+2\left(ab+bc+ca\right)=0\Leftrightarrow ab+bc+ca=\frac{1}{2}\)

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

Mặt khác, ta có \(a^2+b^2+c^2=2\Leftrightarrow\left(a^2+b^2+c^2\right)^2=4\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4\)

\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\Leftrightarrow a^4+b^4+c^4+\frac{1}{2}=4\Leftrightarrow a^4+b^4+c^4=\frac{7}{2}\)

Vậy, ...