Question

Cho \(a+b+c=0\) và \(a^2+b^2+c^2=1\).

Tính giá trị biểu thức: \(a^4+b^4+c^4\)

Answer 1

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

\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)

Ta có: \(a^2+b^2+c^2=1\)

=> \(2\left(ab+bc+ca\right)=-1\)

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

\(\Leftrightarrow\left(ab+bc+ca\right)^2=\dfrac{1}{4}\)

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

Vì a + b + c = 0

=> \(a^2b^2+b^2c^2+c^2a^2=\dfrac{1}{4}\)

Có: \(\left(a^2+b^2+c^2\right)^2=1\)

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

\(\Leftrightarrow a^4+b^4+c^4+2\cdot\dfrac{1}{4}=1\)

\(\Leftrightarrow a^4+b^4+c^4=1-2\cdot\dfrac{1}{4}=1-\dfrac{1}{2}=\dfrac{1}{2}\)