ax+bx=3,ax^2+by^2=7,ax^3+by^3=16,ax^4+by^4=42,求ax^5+bx^5.

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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=

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

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. 2(ax+by-ax)-(ax+by)^2-(by-ax)^2 其中a=-3 b=0.52(ax+by)(by-ax)-(ax+by)^2-(by-ax)^2 2ax+2by=ax+ay+bx+by 怎么计算要写得详细一点 2(ax+by)*(by-ax)-(ax+by)²-(by-ax)²,其中a=-3,x=2分之1 因式分解：(ax-by)^2+(ay+bx)^2 2ax+2ay-bx-by因式分解（ax+by)^2+(bx-ay)^2因式分解 (ax+by)^2+(bx-ay)^2 (ax+by)^2-(ay+bx)^2 (ax+by)^2+(bx-ay)^2 ax+2bx+ay+2by能化简吗? 计算:(-ax by)(bx^2-ay^2). 已知x=2,y=1是二元一次方程组ax+3bx=7,ax-by=1的解,求2a+b的值实数a,b,x,满足ax+by=3,ax^2+by^2=7,ax^3+by^3=16,ax^4+by^4=42,ax^5+by^5=? 若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^5+by^5的值若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^5+by^5的值 (ax+by)^2(ay+bx)^2+2(ax+by)(ay+bx)