Ta có : a+b+c=3⇒(a+b+c)2=9
⇒a2+b2+c2+2(ab+bc+ca)=9
⇒a2+b2+c2=9−2(ab+bc+ca)=9−2.3=3
⇒a2+b2+c2=ab+bc+ca
⇔a2+b2+c2−ab−bc−ca=0
⇔2a2+2b2+2c2−2ab−2bc−2ca=0
⇔(a2−2ab+b2)+(b2−2bc+c2)+(c2−2ca+a2)=0
⇔(a−b)2+(b−c)2+(c−a)2=0
⇔a=b=c
Mà a+b+c=3⇒a=b=c=1
⇒M=(1−1−1)2018+(1−1−1)2019+(1−1−1)2020=1−1+1=1
Vậy M=1