Cho a, b>0 a^2000+b^2000=a^2001+b^2001=a^2002+b^2002 .Tính a^2011+b^2011

Có $a^{2000}+b^{2000}= a^{2001}+b^{2001} = a^{2002}+b^{2002}$

$\to a^{2000}+b^{2000} + a^{2002}+b^{2002} = 2.(a^{2001}+b^{2001})$

$\to a^{2002} - a^{2001} - a^{2001} + a^{2000} + b^{2002} - b^{2001} - b^{2001} + b^{2000}  = 0$

$\to a^{2001}.(a-1) - a^{2000}.(a-1) + b^{2001}.(b-1) - b^{2000}.(b-1)  = 0 $

$\to (a-1).(a^{2001}-a^{2000}) + (b-1).(b^{2001}-b^{2000} ) = 0 $

$\to (a-1)^2.a^{2000} + (b-1)^2.b^{2000} = 0 $

Dấu "=" xảy ra $⇔(a-1)^2.a^{2000}=0;  (b-1)^2.b^{2000} = 0 $

Mà $a,b>0$

$\to a=b=1$

Khi đó $a^{2011}+b^{2011} = 1+1=2$