Question

tinh a^4+b^4+c^4 biết: a+b+c=0 và a^2+b^2+c^2=2

Answer 1

Ta có :  \(a+b+c=0\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\Leftrightarrow ab+bc+ac=-1\)​

​\(\left(ab+bc+ac\right)^2=1\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=1\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=1\)​

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

Answer 2

Ta có:

a+b+c=0 => (a+b+c)²=0 => a²+b²+c² = -2(ab+bc+ac)

=> a⁴+b⁴+c⁴ + 2(a²b²+b²c² + a²c²) = 4(a²b²+b²c² + a²c²)+8(ab²c + abc² + a²bc)

=> a⁴+b⁴+c⁴ =2(a²b²+b²c² + a²c²) + 8abc(a+b+c)

=> a⁴+b⁴+c⁴ =2(a²b²+b²c² + a²c²)

Mặt khác, vì

a²+b²+c² = -2(ab+bc+ac)=2

=> ab +bc+ac = -1

=>a²b²+b²c² + a²c²+2(ab²c + abc² + a²bc) = 1

=> a²b²+b²c² + a²c² = 1

=> a⁴+b⁴+c⁴  = 1* 2 =2