ax+bx=3,ax^2+by^2=7,ax^3+by^3=16,ax^4+by^4=42,求ax^5+bx^5.
来源:学生作业帮助网 编辑:六六作业网 时间:2024/12/27 23:14:42
ax+bx=3,ax^2+by^2=7,ax^3+by^3=16,ax^4+by^4=42,求ax^5+bx^5.
ax+bx=3,ax^2+by^2=7,ax^3+by^3=16,ax^4+by^4=42,求ax^5+bx^5.
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=7
(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