cho a+b+c=0 và a2+b2+c2=2017

tính A=a4+b4+c4

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

\(\Rightarrow2\left(ab+bc+ca\right)=-2017\)

\(\Rightarrow ab+bc+ca=\dfrac{-2017}{2}\)

\(\Rightarrow\left(ab+bc+ca\right)^2=1017072,25\)

\(\Rightarrow\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2abc\left(a+b+c\right)=1017072,25\)

\(\Rightarrow a^2b^2+b^2c^2+c^2a^2=1017072,25\)

Ta có: \(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=2017^2\)

\(\Rightarrow a^4+b^4+c^4+2.1017072,25=2017^2\)

\(\Rightarrow a^4+b^4+c^4=3051216,75\)

Ta có : \(a^3+b^3=c\left(3ab-c^2\right)\)

\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-bc-ca+c^2\right)-3ab\left(a+b+c\right)=0\)

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

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\) ( Vì \(a+b+c=3\) )

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

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

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

\(\Leftrightarrow a=b=c\)

Mà : \(a+b+c=3\Rightarrow a=b=c=1\)

\(\Rightarrow A=675\left(1^{2018}+1^{2018}+1^{2018}\right)+1=675.3+1=2026\)