A [解析] 设f(x)=x3+2x2-5x-6,则 f(x)=(x+1)(x-x2)(x-x3) =x3+(1-x2-x3)x2+(x2x3-x2-x3)x+x2x3 =x3+2x2-5x-6 对比同次项系数,得\(1-x_2-x_1=2x_2x_3-x_2-x_3=-5. 即 \(x_2+x_3=-1x_2x_3=-6. 所以 1/(x_2)+1/(x_3)=(x_2+x_...
解方程:x3+2x2-5x-6=0. 答案 x3+2x2-5x-6+x-x=0, x3+2x2+x-6x-6=0 x(x2+2x+1)-6(x+1)=0 x(x+1)2-6(x+1)=0 (x+1)⎡⎢⎣⎤⎥⎦x(x+1)-6=0 (x+1)(x2+x-6)=0 (x+1)(x+3)(x-2)=0 解得:x1=-1,x2=-3,x3=2.故答案为:x1=-1,...
解答解:∵当x=-1时,x3+2x2-5x-6=0, ∴可设x3+2x2-5x-6=(x+1)(x2+ax-6), 展开(x+1)(x2+ax-6)=x3+(a+1)x2+(a-6)x-6, ∴{a+1=2a−6=−5{a+1=2a−6=−5,解得a=1. ∴x3+2x2-5x-6=(x+1)(x2+x-6)=(x+1)(x-2)(x+3). ...
所以a的倒数加b的倒数等于a+b除以ab，即为六分之一。x³+2x²-5x-6=0x³+x²+x²-5x-6=0x²(x+1)+(x+1)(x-6)=0(x+1)(x²+x-6)=0(x+1)(x+3)(x-2)=0所以：x=-1,或x=-3或x=2可得另两个根的倒数和为：-1/3+1/2=1/6
E.-2或-3 点击查看答案&解析手机看题 单项选择题 已知方程x 3 +2x 2 -5x-6=0的根为x 1 =-1,x 2 ,x 3 ,则 A. B. C. D. E. 点击查看答案&解析手机看题 单项选择题 已知方程x 3 -2x 2 -2x+1=0有三个根x 1 ,x 2 ,x 3 ,其中x 1 =-1,则|x 2 -x 3 |等于___ ...
(x^3 + x^2) + (x^2 - 5x - 6) = x^2*(x+1) + (x+1)(x-6) = (x+1)(x^2 + x-6) (x+1)(x+3)(x-2) = 0 所以,它的解有:x = -1,x = -3,x = 2结果一 题目 x³+2x²-5x-6=0 答案 (x^3 + x^2) + (x^2 - 5x - 6) = x^2*(x+1) + (x+1)...
结果一 题目 【题目】因式分解: x^3+2x^2-5x-6 答案 【解析】原式=(x^3+x^2)+(x^2+x)-(6x+6) =x^2(x+1)+x(x+1)-6(x+1)=(x+1)(x^2+x-6) =(x+1)(x-2)(x+3)相关推荐 1【题目】因式分解: x^3+2x^2-5x-6
题目说 当x=-1时,x^3+2x^2-5x-6=0,这句话就已经表示此多项式必定有x+1=0这个因式了 分解因式x^3+2x^2-5x-6
因为x=-1时,x3+2x2-5x-6=0,所以x^3+2x^2-5x-6有一个因式为x+1,设x^3+2x^2-5x-6=(x+1)(x^2+ax-6)=x^3+(a+1)x^2+(a-6)x-6,比较系数，得，a+1=2,解得a=1,所以x^3+2x^2-5x-6=(x+1)(x^2+x-6)=(x+1)(x+3)(x-2)x...
解答解:(1)x2-5x-6=0, (x+1)(x-6)=0, ∴x+1=0,x-6=0, x1=-1,x2=6. (2)2(x-3)=3x(x-3), 2(x-3)-3x(x-3)=0 (x-3)(2-3x)=0, ∴x-3=0,2-3x=0, ∴x1=3,x2=2323; (3)∵x2-2x-5=0, ∴x2-2x=5, ...