Question

问几道数学题

1.在复数集内因式分解x^3-y^3=
2.2x^3-4x^2+bx-2=0有一个实根为1,另外两个根为虚根，则此方程的解是
3.方程x^2-x+1=0的两个根为a,b,则a^10+b^10=
4.在数列an中,Sn=n^2/(2n+2),求liman

Answer 1

1、x^3-y^3=（x-y）(x^2+xy+y^2)=(x-y)[x-(-1+√3i)y] [x-(-1-√3i)y]
=(x-y)[x+(1-√3i)y] [x+(1+√3i)y]
2、2-4+b-2=0,b=4,
原方程为：.2x^3-4x^2+4x-2=0，
提取x-1公因式，2（x-1）(x^2-x+1)=0,
x2、3=(1±√3i)/2.
3、由2题解出a=(1+√3i)/2，b=(1-√3i)/2
化成三角式，a=cos（π/3）+isin(π/3),
b=cos（-π/3）+isin(-π/3),
根据棣美弗定理，
a^10+b^10=[ cos（π/3）+isin(π/3)]^10+[ cos（-π/3）+isin(-π/3)]^10
= cos（10π/3）+isin(10π/3)+ cos（-10π/3）+isin(-10π/3)
= cos（10π/3）+isin(10π/3)+ cos（10π/3）-isin(10π/3)
= 2cos（10π/3）
=2cos(2π+π+π/3)
=-2cos(π/3)
=-2*(1/2)
=-1.
4、Sn=n^2/(2n+2),
a(n)=S(n)-S(n-1)= n^2/(2n+2)-(n-1)^2/[2(n-1)+2]
=(n^2+n-1)/[2n(n+1)],
lim[n→∞]a(n)= lim[n→∞] (n^2+n-1)/[2n(n+1)]
= lim[n→∞](1+1/n-1/n^2)/[2(1+1/n)]
=1/2.