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# 一道难题，回答对的送分。

jiangyang6 2002-07-27 09:01:08

a*x^3+b*x^2+c*x+d=0;

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8 条回复

zhoukun666 2002-07-30
up

dcyu 2002-07-30

Reduce[a*x^3+b*x^2+c*x+d==0,x]

(i) 当 a!=0 时

x1=-b/(3*a)-2^(1/3)*(-b^2+3*a*c)/3/a/t + t/3/2^(1/3)/a (实数解)
x2=-b/(3*a)+(1+i*sqrt(3))*(-b^2+3*a*c)/3/a/t/2^(1/3) - (1-i*sqrt(3))*t/6/a/2^(1/3)
x3=-b/(3*a)+(1-i*sqrt(3))*(-b^2+3*a*c)/3/a/t/2^(1/3) - (1+i*sqrt(3))*t/6/a/2^(1/3)
(ii) 当 a==0,b!=0 时
x1=(-c-sqrt(c^2-4*b*d))/2/b
x2=(-c+sqrt(c^2-4*b*d))/2/b
(iii)当 a==0,b==0,c!=0 时
x=-d/c
(iv)当 a==0,b==0,c==0 时

Bonny_lj 2002-07-29

mmmcd 2002-07-28

c_and_pascal 2002-07-28

(1) a*x = b
x = b / a
(2) a*x*x + b*x + c = 0
x = (-b+/-sqrt(b^2-4*a*c))/(2*a)
(3)
(a)
x*x*x - 1 = 0
x1 = 1, x2 = w = (-1 + i*sqrt(3)), x3 = w*w = (-1 -i*sqrt(3))/2

(b)
a*x*x*x + b*x*x + c*x + d = 0
x*x*x + p*x + q = 0 (卡尔丹公式）
x1 = (-q/2 + ((q/2)^2 + (p/3)^3)^(1/2))^(1/3)
+ (-q/2 - ((q/2)^2 + (p/3)^3)^(1/2))^(1/3)
x2 = w*(-q/2 + ((q/2)^2 + (p/3)^3)^(1/2))^(1/3)
+ w*w*(-q/2 - ((q/2)^2 + (p/3)^3)^(1/2))^(1/3)
x1 = w*w*(-q/2 + ((q/2)^2 + (p/3)^3)^(1/2))^(1/3)
+ w*(-q/2 - ((q/2)^2 + (p/3)^3)^(1/2))^(1/3)

(c)
a*x*x*x + b*x*x + c*x + d = 0

(4)
(a) a*x*x*x*x + c*x*x + e = 0
y = x*x : a*y*y + c*y + e = 0

(b) a*x*x*x*x = b&x&x&x + c*x*x + b*x + a = 0
y= x + 1/x :
x1,2,3,4 = (y +- sqrt(y*y - 4)) / 2
y = (-b +-sqrt(b*b-4*a*c+8*a*a))/(2*a)

(c) a=1 : x^4 + b*x^3 + c*x^2 + d*x + e =0
x^2 + (b + sqrt(8*y + b^2 - 4*c)*x/2 + (y+(b*y-d)/(sqrt(8*y + b^2 - 4*c)) = 0
y是:8*y^3 - 4*c*y^2 + (2*b*d-8*e)*y + e*(4*c-b^2) - d^2 = 0

(5) 阿贝尔定理

(6) 整系数求有理解

Renny 2002-07-28

x^3-6*x-6=0

x^3-3*a*b*x-(a^3+b^3)=0
3*a*b=6,a^3+b^3=6
27*a^3*b*3=216, a^3+b^3=6
a^3*b^3=8,a^3+b^3=6

a^3=4,b^3=2 因为是对称的，所以我们不用考虑
a^3=2,b^3=4

x=4^(1/3)+2^(1/3)

Lawrence444 2002-07-27

Lawrence444 2002-07-27

(x-x1)(x-x2)(x-x3)...(x-xn)=0

x^n+y1*x^(n-1)+y2*x^(n-2)+...=0

x^n+(b/a)*x^(n-1)+(c/a)*x^(n-2)+...=0

y1=b/a
y2=c/a
...

3.2w+