2013年国际足联排名:初二奥赛题……高手进!急!

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1 若x+y=a+b 且x^2+y^2=a^2+b^2 ,求证:
x^2006+y^2006=a^2006+b^2006
2 设a,b为实数,求a^2+ab+b^2-a-2b的最小值.

x^2+y^2=(x+y)^2-2xy=(a+b)^2-2xy=a^2+b^2
xy=1/2[(a+b)^2-(a^2+b^2)]=ab

x^3+y^3=(x+y)(x^2+y^2)-xy(x+y)=a^3+b^3

设x^n+y^n=a^n+b^n,x^(n-1)+y^(n-1)=a^n-1+b^n-1.
x^(n+1)+y^(n+1)
=(x+y)(x^n+y^n)-(xy)[x^(n-1)+y^(n-1)]=a^n+1+b^n+1
因为n=2成立,所以x^2006+y^2006=a^2006+b^2006

用归纳法做!

2。a^2+ab+b^2-a-2b
=()^2+()^2+.....

x^2+y^2=(x+y)^2-2xy=(a+b)^2-2xy=a^2+b^2
xy=1/2[(a+b)^2-(a^2+b^2)]=ab
x^3+y^3=(x+y)(x^2+y^2)-xy(x+y)=a^3+b^3
设x^n+y^n=a^n+b^n,x^(n-1)+y^(n-1)=a^n-1+b^n-1.
x^(n+1)+y^(n+1)
=(x+y)(x^n+y^n)-(xy)[x^(n-1)+y^(n-1)]=a^n+1+b^n+1
因为n=2成立,所以x^2006+y^2006=a^2006+b^2006
用归纳法做!
2。a^2+ab+b^2-a-2b
=()^2+()^2+.....

x^2+y^2=(x+y)^2-2xy=(a+b)^2-2xy=a^2+b^2
xy=1/2[(a+b)^2-(a^2+b^2)]=ab
x^3+y^3=(x+y)(x^2+y^2)-xy(x+y)=a^3+b^3
设x^n+y^n=a^n+b^n,x^(n-1)+y^(n-1)=a^n-1+b^n-1.
x^(n+1)+y^(n+1)
=(x+y)(x^n+y^n)-(xy)[x^(n-1)+y^(n-1)]=a^n+1+b^n+1
因为n=2成立,所以x^2006+y^2006=a^2006+b^2006
用归纳法做!
2。a^2+ab+b^2-a-2b
=()^2+()^2+.....

1、x+y=a+b =>x-a=b-y
x^2+y^2=a^2+b^2 =>x^2-a^2=b^2-y^2 =>(x+a)(x-a)=(b-y)(b+y)
所以x+a=b+y.结合x-a=b-y易得x=b,y=a 显然x^2006+y^2006=a^2006+b^2006

2、a^2+ab+b^2-a-2b=(2a^2+2ab+2b^2-2a-4b)/2
=((a+b)^2+(a-0.5)^2+(b-2)^2-17/4)/2
分析:(a+b)^2,(a-0.5)^2,(b-2)^2正定,所以a=0.5,b=2时最小。故a^2+ab+b^2-a-2b的最小值=(2.5^2-17/4)/2=1

1.
x+y=a+b
x-a=b-y

x^2+y^2=a^2+b^2
x^2-a^2=b^2-y^2
(x+a)(x-a)=(b+y)(b-y)
讨论:
(1) x-a=b-y=0,则x=a,b=y
x^2006+y^2006=a^2006+b^2006
(2)x-a=b-y≠0,则x+a=b+y
把x-a=b-y和x+a=b+y相加
x=b,y=a
x^2006+y^2006=a^2006+b^2006

2.
a^2+ab+b^2-a-2b
=a^2+(b-1)a+b^2-2b
=[a+(b-1)/2]^2-(b-1)^2/4+b^2-2b
=[a+(b-1)/2]^2+3b^2/4-3b/2-1/4
=[a+(b-1)/2]^2+3(b^2-2b)/4-1/4
=[a+(b-1)/2]^2+[3(b-1)^2]/4-1
当b-1=0,a+(b-1)/2=0时,即b=1,a=0
a^2+ab+b^2-a-2b有最小值-1