Question

[Bui thi hai ha]

Cho a+b+c=0 và a² + b² + c² =1 . Tính giá trị biểu thức M=a⁴+b⁴+c⁴.

Answer 1

[Minh Anh]

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

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

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

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

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

\(\Rightarrow2\left(ab+bc+ac\right)=-1\)

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

\(\Rightarrow\left(ab+bc+ac\right)^2=\left(-\frac{1}{2}\right)^2=\frac{1}{4}\)

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

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

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

Vậy: \(a^4+b^4+c^4=1-2\left(a^2b^2+b^2c^2+a^2c^2\right)\)

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

Answer 2

[Đặng Tuấn Anh]

M = 1/2