cho 2 so thuc a va b thoa man a^3-3a^2+5a-17=0 va b^3-3b^2+5b+11=0

(1)⇔(a−1)^3+2(a−1)−14=0(1)
⇔(a−1)^3+2(a−1)−14=0
 
(2)⇔(b−1)^3+2(b−1)+14=0(2)
⇔(b−1)^3+2(b−1)+14=0
 
Cộng theo vế,đc:
 
(a+b−2)[(a−1)2−(a−1)(b−1)+(b−1)2]+2(a+b−2)=0(a+b−2)[(a−1)2−(a−1)(b−1)+(b−1)2]+2(a+b−2)=0
 
⇔a+b=2