Question

[ONLINE SWORD ART]

Cho 3 số a,b,c thỏa mãn a+b+c=0; a^2 + b^2+c^2=14

tính giá trị của biểu thức: M=a^4 +b^4 + c^4

Answer 1

[FLT24]

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

\(\Leftrightarrow14+2\left(ab+bc+ac\right)=0\Leftrightarrow ab+bc+ac=-7\)

Suy ra : \(\left(ab+bc+ac\right)^2=49\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=49\)

\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2=49\)

\(a^2+b^2+c^2=14\Leftrightarrow\left(a^2+b^2+c^2\right)^2=196\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=196\)

\(\Leftrightarrow a^4+b^4+c^4+2.49=256\)  \(\Leftrightarrow a^4+b^4+c^4=98\)

Vậy ... 

Answer 2

[Trần Tuấn Hoàng]

\(a+b+c=0\)

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

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

\(\Leftrightarrow2ab+2bc+2ca=-14\)

\(\Leftrightarrow ab+bc+ca=-7\)

\(\Rightarrow\left(ab+bc+ca\right)^2=49\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2ab^2c+2abc^2+2a^2bc=49\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=49\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=49\).

\(a^2+b^2+c^2=14\)

\(\Rightarrow\left(a^2+b^2+c^2\right)^2=14^2=196\)

\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=196\)

\(\Leftrightarrow a^4+b^4+c^4+2.49=196\)

\(\Leftrightarrow a^4+b^4+c^4=98\)