Question

[Hoàng Phúc]

Cho a,b dương và a^2000+b^2000=a^2001+b^2001=a^2002+b^2002

Tính a^2011 + b^2011

Answer 1

[ngonhuminh]

\(a^{2000}+b^{2000}=a.a^{2000}+b.b^{2000}=a^2.a^{2000}+b^2.b^{2000}\)

a=b={0,1} là nghiệm 

xét a,b \(\ne\left\{0,1\right\}\)

\(\left(1-a\right).a^{2000}=\left(b-1\right).b^{2000}\Leftrightarrow\frac{1-a}{b-1}=\left(\frac{b}{a}\right)^{2000}\)(1)

\(\left(1-a^2\right).a^{2000}=\left(b^2-1\right).b^{2000}\Rightarrow\frac{1-a^2}{b^2-1}=\left(\frac{b}{a}\right)^{2000}\)(2)

(1)&(2)=>\(\frac{1-a}{b-1}=\frac{1-a^2}{b^2-1}\Rightarrow\left(1-a\right)\left(b+1\right)=\left(1-a\right)\left(1+a\right)\Rightarrow a=b\)

Thay vào phương trình đầu: => a=b={0,1) a, b dương => a=b=1

a^20011+b^20011=2

Answer 2

[Cold Wind]

\(a^{2000}+b^{2000}=a^{2001}+b^{2001}=a^{2002}+b^{2002}\)

\(\Leftrightarrow a^{2000}+b^{2000}=a\cdot a^{2000}+b\cdot b^{2000}=a^2\cdot a^{2000}+b^2\cdot b^{2000}\)

Mà a,b >0 

\(\Rightarrow\hept{\begin{cases}a=a^2=1\\b=b^2=1\end{cases}\Rightarrow a=b=1}\)

Vậy \(a^{2011}+b^{2011}=1+1=2\)

True or False??!?