实数a,b,x,满足ax+by=3,ax^2+by^2=7,ax^3+by^3=16,ax^4+by^4=42,ax^5
1个回答

因为ax2+by2=7

则ax2=7-by2,by2=7-ax2

ax3=7x-bxy2,by3=7y-ax2y

ax3+by3=7(x+y)-xy(by+ax)=16

即7(x+y)-3xy=16

又因为ax3+by3=16

则ax3=16-by3,by3=16-ax3

ax4=16x-bxy3,by4=16y-ax3y

ax4+by4=16(x+y)-xy(by2+ax2)=42

即16(x+y)-7xy=42

由两式组成方程组:7(x+y)-3xy=16

16(x+y)-7xy=42

解得x+y=-14,xy=-38

又因为 ax4+by4=42

ax4=42-by4,by4=42-ax4

ax5=42x-bxy4,by5=42y-ax4y

ax5+by5=42(x+y)-xy(by3+ax3)

=42*(-14)-16*(-38)

=-588+608

=20