Question

Cho \(a^2+b^2+c^2=a^3+b^3+c^3=1\). Tính \(S=a^2+b^{2012}+c^{2013}\).

Answer 1

Ta có: a2+b2+c2=1a2+b2+c2=1

⇒|a|≤1 |b|≤1 |c|≤1⇒|a|≤1 |b|≤1 |c|≤1

Ta lại có:

a3+b3+c3=a2+b2+c2a3+b3+c3=a2+b2+c2

⇔a2(1−a)+b2(1−b)+c2(1−c)=0⇔a2(1−a)+b2(1−b)+c2(1−c)=0

Vì 1−a≥0 1−b≥0 1−c≥0 1−a ≥0 1−b≥0 1−c≥0

⇒a2(1−a)+b2(1−b)+c2(1−c)≥0⇒a2(1−a)+b2(1−b)+c2(1−c)≥0

Dấu = xảy ra khi: (a,b,c)=(1,0,0;0,1,0;0,0,1)(a,b,c)=(1,0,0;0,1,0;0,0,1)

⇒S=1