(ax^2+by^2)(x+y)=7(x+y)
ax^3+by^3+ax^2y+bxy^2=7(x+y)
ax^3+by^3+xy(ax+by)=7(x+y)
以ax+by=3,ax^3+by^3=16代入上式
16+3xy=7(x+y)
ax^3+by^3=16
(ax^3+by^3)(x+y)=16(x+y)
ax^4+by^4+xy(ax^2+by^2)=16(x+y)
以ax^4+by^4=42,ax^2+by^2=7代入上式
42+7xy=16(x+y)
综上所述
7(x+y)-3xy=16
16(x+y)-7xy=42
将(x+y)和xy看成整体的两个未知数。解方程组
7a-3b=16
16a-7b=42
49a-21b=112
48a-21b=126
a=-14
7*(-14)-3b=16
b=-38
因此
x+y=-14
xy=-38
ax^4+by^4=42
两端同时乘以(x+y)
(ax^4+by^4)(x+y)=42(x+y)
ax^5+by^5+(ax^3+by^3)xy=42(x+y)
以ax^3+by^3=16代入上式，得到
ax^5+by^5=42(x+y)-16xy
=42*(-14)-16*(-38)
=-588+608
=20
ax+bx=3,ax^2+by^2=7,ax^3+by^3=16,ax^4+by^4=42,求ax^5+bx^5.
(ax^2+by^2)(x+y)=7(x+y)ax^3+by^3+ax^2y+bxy^2=7(x+y)ax^3+by^3+xy(ax+by)=7(x+y)以ax+by=3,ax^3+by^3=16代入上式 16+3xy=7(x+y)ax^3+by^3=16 (ax^3+by^3)(x+y)=16(x+y)ax^4+by^4+xy(ax^2+by^2)=16(x+y)以ax^4+by^4=42,ax^2+by^...
设a.b.x.y满足ax+by=3,ax^2+by^2=7,ax^3+by^3=16,ax^4+by^4=42,求ax...
答案是20。解：因为(ax^2+by^2)*(x+y)=(ax^3+by^3)+xy(ax+by),所以有7(x+y)=16+3xy...(1)因为(ax^3+by^3)*(x+y)=(ax^4+by^4)+xy(ax^2+by^2),所以有16(x+y)=42+7xy...(2)因为(ax^4+by^4)*(x+y)=(ax^5+by^5)+xy(ax^3+by^3),所以有42(x+y)...
若abxy满足ax+by=3,ax^2+by^2=7,ax^3+by^3=16,ax^4+by^4=42,求ax^5...
∵ (ax^2+by^2)·(x+y)=(ax^3+by^3)+xy(ax+by),∴ 7(x+y)=16+3xy ………① ∵ (ax^3+by^3)·(x+y)=(ax^4+by^4)+xy(ax^2+by^2),∴ 16(x+y)=42+7xy ………②∵ (ax^4+by^4)·(x+y)=(ax^5+by^5)+xy(ax^3+by^3),∴ 42(x+y)=(ax^5+by...
...ax+by=3,ax^2+by^2=7,ax^3+by^3=16,ax^4+by^4=42,ax^5+by^5=...
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+...
设a,b,x,y满足 ax+by=3,ax^2+by^2=7,ax^3+by^3=16,ax^4+by^4=42,求...
则 7(x+y)=16+3xy...(1)(ax^3+by^3)*(x+y)=(ax^4+by^4)+xy(ax^2+by^2)，则 16(x+y)=42+7xy...(2)(ax^4+by^4)*(x+y)=(ax^5+by^5)+xy(ax^3+by^3)，则 42(x+y)=(ax^5+by^5)+16xy...(3)由(1)和(2) 可得xy=-38，x+y=-14，代入(3)得 ...
...=7,ax²+by³=16,ax^4+by^4=42,求ax^5+by^5的值
即：7(x+y)=16+3xy ① (ax³+by³)(x+y)=ax^4+by^4+xy(ax²+by²)即：16(x+y)=42+7xy ② 解①、②联立的方程组得：x+y=-14 xy=-38 (ax^4+by^4)(x+y)=ax^5+by^5+xy(ax³+by³)42*(-14)=ax^5+by^5-38*16 ax^5+...
已知:ax+by=3 ax*2+by*2=7 ax*3+by*3=16 ax*4+by*4=42 ax ax
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...
已知:ax+by=3 ax2+by2=7 ax3+by3=16 ax4+by4=42 求:ax5+by5
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 ...
已知ax+by=3,ax^2+by^2=6,ax^3+by^3=18,ax^4+by^4=72,求ax^5+by^5的...
ax^1+by^1=3, 3 ax^2+by^2=6, 2*3：幂×上一个结果 ax^3+by^3=18, 3*6：幂×上一个结果 ax^4+by^4=72, 4*18：幂×上一个结果 求ax^5+by^5=360 5*72：幂×上一个结果
已知:ax+by=3 ax2+by2=7 ax3+by3=16 ax4+by4=42 求:ax5+by5
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 ...
