初二奥赛题……高手进！急！

来源：百度知道 编辑：UC知道 时间：2024/05/31 14:34:48
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