作业帮 > 数学 > 作业

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 编辑:搜狗做题网作业帮 分类:数学作业 时间:2024/04/30 18:54:21
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