因式分解练习ax+by=3ax2+by2=7ax3+by3=16ax4+by4=42求ax5+by5=?(x+y)(y+z)(x+z)+xyz6x2-15y2-6z2-xy-5xz+21yz

Question

因式分解练习
ax+by=3 ax2+by2=7 ax3+by3=16 ax4+by4=42 求ax5+by5=?
(x+y)(y+z)(x+z)+xyz
6x2-15y2-6z2-xy-5xz+21yz

Answer 1

第一题：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
第二题：(x+y)(y+z)(x+z)+xyz=x+y+z-xyz+xyz=x+y+z
第三题：6x2-15y2-6z2-xy-5xz+21yz =(2 x + 3 y - 3 z) (3 x - 5 y + 2 z)