Question

Cho a+b+c=0, a²⁺b²⁺c²=1

chứng minh rằng a⁴⁺b⁴⁺c⁴=1^​/2

Answer 1

Có: \(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+ac+bc\right)=0\)

\(\Rightarrow a^2+b^2+c^2=-2\left(ab+ac+bc\right)\)

Theo bài ra: \(a^2+b^2+b^2=1\)

\(\Rightarrow-2\left(ab+ac+bc\right)=1\Rightarrow ab+ac+bc=-\frac{1}{2}\)

Lại có: \(a^2+b^2+c^2=1\Rightarrow\left(a^2+b^2+c^2\right)^2=1\)

\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2=1\)

\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+a^2c^2+b^2c^2\right)=1\)

Mà: \(2\left(a^2b^2+a^2c^2+b^2c^2\right)=2\left(ab+ac+bc\right)^2=2.\left(-\frac{1}{2}\right)^2=\frac{1}{2}\)

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

Vậy:...