x^n-y^n因式分解证明

还是老样子,极限的定义,无限分有限+无限lim(x(n+1)-x(n))/(y(n+1)-yn)存在设lim(x(n+1)-x(n))/(y(n+1)-yn)=a对于任意e>0,存在N使得,对n>N有

答：（x-y）的2N次方-（x-y）的2n+1次方+（y-x）的2n-1次方=(x-y)^(2n)-(x-y)^(2n+1)-(x-y)^(2n-1)=[(x-y)-(x-y)^2-1]*(x-y)^

5m(x-y)²+10n(y-x)³=5m(x-y)²-10n(x-y)³=5(x-y)²[m-2n(x-y)]=5(x-y)²(m-2nx

m²+2m(m-n)+(m-n)²=(m+m-n)²=(2m-n)²16-8（x-y）+（x-y）²=[4-(x+y)]²=(4-x-y)&

您好：5x平方y(m-n)平方-10xy平方(n-m)三次方=5x平方y(m-n)平方+10xy平方(m-n)三次方=5xy(m-n)²（x-2m+2n）如果本题有什么不明白可以追问,如果满

x²-2xy+y²-2x+2y+1=(x-y)²-2(x-y)+1=(x-y-1)²-x^(n+1)+2x^n-x^(n-1)=-x^(n-1)*x^2+2x^

-3x^(n+2)+18x^(n+1)y-27x^ny^2=-3x^n(x^2-6xy+9y^2)=-3x^n(x-3y)^2

当n为正偶数时,(y-x)^n=（y-x）^n当n为正奇数时,(y-x)^n=（x-y）^n

x^n-y^n=(x-y)(x^n-1+x^n-2y+x^n-3y^2+.+xy^n-2+y^n-1)

3a(x-y)+9(y-x)=3a(x-y)-9(x-y)提取公因式3(x-y)=3(x-y)(a-3)(2m-3n)²-2m+3n=(2m-3n)²-(2m-3n)提取公因式(2

(2m+3n)(2m-n)-4n(2m-n)=(2m-n)(2m+3n-4n)=(2m-n)(2m-n)=(2m-n)^2(x+y)^2(x-y)+(x+y)(y-x)^2=(x+y)(x-y)(x+

x^(n+1)y^n-x^ny^(n+1)=x^n*y^n*(x-y)=(xy)^n*(x-y)

解(m-n)²-2m+2n=(m-n)²-2(m-n)=(m-n)(m-n-2)=(x-y)(x+y)-(x+y)²=(x+y)(x-y-x-y)=-2y(x+y)

4(x+2y)^2-9(x-y)^2=[2(x+2y)+3(x-y)][2(x+2y)-3(x-y)]（平方差公式）=-(5x+y)(x-7y)x^(n+3)-x^(n-1)=x^(n-1)(x^4-

费马最后定理:当n是一个整数且n>2时,方程x^n+y^n=z^n无正整数x,y,z的解Euler证明的n=3,4的情形,对于该问题,只需证明n为素数的情形.谷山-志村定理"所有Q上的椭圆曲线是模的"

答：(x-y)^2(m-n)+(x+y)^2(n-m)=(m-n)*[(x-y)^2-(x+y)^2]=(m-n)*(x-y+x+y)*(x-y-x-y)=(m-n)*(2x)*(-2y)=-4(m-

(m-n)²+2（n-m）=_______=(m-n)²-2(m-n)=(m-n-2)(m-n)x²-y²+6y-9=______=x²-(y

分两种情况：1.n为奇数时a(x-y)的n次方-3b(y-x)的n+1次方+2c(y-x)的n+2=a(x-y)^n－3b(x－y)^(n＋1)－2c(x－y)^(n＋2)=(x-y)^n*[a－3b

(x+y)(5m+3n)²-(x+y)(m-n)²=(x+y)[(5m+3n)²-(m-n)²]=(x+y)[(5m+3n)+(m-n)][(5m+3n)-(m

1.当n=1时原式=x^2-y^2=(x-y)(x+y)能被x+y整除故命题成立2.假设n=k时命题成立,即x^(2k)-y^(2k)能被x+y整除当n=k+1时x^(2k+2)-y^(2k+2)=x