数学难题已知ax+by=3,ax^2+by^2=7ax^3+by^3=16,ax^4+by^4=42求：ax^5=by^5的值谢谢啦~~~是ax^5+by^5的值是多少

Question

数学难题
已知ax+by=3,ax^2+by^2=7
ax^3+by^3=16,ax^4+by^4=42
求：ax^5=by^5的值
谢谢啦~~~
是ax^5+by^5的值是多少

Answer 1

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