Cho a, b>0 a^2000+b^2000=a^2001+b^2001=a^2002+b^2002 .Tính a^2011+b^2011 07/08/2021 Bởi Alice Cho a, b>0 a^2000+b^2000=a^2001+b^2001=a^2002+b^2002 .Tính a^2011+b^2011
`a^2000 + b ^2000` ⇒ `a^2000(a – 1) + b^2000(b – 1) = 0 (1)` ⇒ `a^2001(a – 2) + b^2001(b – 2) = 0 (2)` `(2) – (1) = a^2001(a – 1) + b^2001(b – 1) ` `text{a = b = 1}` ⇔`a^2011+b^2011 = 1 + 1 = 2` Bình luận
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 $a^{2011}+b^{2011} = 1+1=2$ Bình luận
`a^2000 + b ^2000`
⇒ `a^2000(a – 1) + b^2000(b – 1) = 0 (1)`
⇒ `a^2001(a – 2) + b^2001(b – 2) = 0 (2)`
`(2) – (1) = a^2001(a – 1) + b^2001(b – 1) `
`text{a = b = 1}`
⇔`a^2011+b^2011 = 1 + 1 = 2`
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 $a^{2011}+b^{2011} = 1+1=2$