Question

[Vũ Đăng Thành]

tính giá trị của BT \(a^4+b^{4^{ }}+c^4+\dfrac{1}{4}\) biết a+b+c = 0 và \(a^2+b^2+c^2=1\)

Answer 1

[Nguyễn Lê Phước Thịnh]

Ta có: a+b+c=0

nên \(\left(a+b+c\right)^2=0\)

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

\(\Leftrightarrow2ab+2ac+2bc=-1\)

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

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

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

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

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

Ta có: \(a^2+b^2+c^2=1\)

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

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

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

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

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

\(\Leftrightarrow a^4+b^4+c^4+\dfrac{1}{4}=\dfrac{1}{2}+\dfrac{1}{4}=\dfrac{2}{4}+\dfrac{1}{4}=\dfrac{3}{4}\)

Vậy: \(a^4+b^4+c^4+\dfrac{1}{4}=\dfrac{3}{4}\)