卡当公式推导过程在百度百科上看见:“由一元三次方程的完整式X3+a1X2+a2X+a3=0 (1)和缺项式X3+pX+q=0 (2)可知,欲将式 (1)转换为式 (2),需令y=X-a1/3代入式 (1),得(X-a1/3)3+a1(X-a1/3)2+…=0,”其
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![卡当公式推导过程在百度百科上看见:“由一元三次方程的完整式X3+a1X2+a2X+a3=0 (1)和缺项式X3+pX+q=0 (2)可知,欲将式 (1)转换为式 (2),需令y=X-a1/3代入式 (1),得(X-a1/3)3+a1(X-a1/3)2+…=0,”其](/uploads/image/z/1347313-49-3.jpg?t=%E5%8D%A1%E5%BD%93%E5%85%AC%E5%BC%8F%E6%8E%A8%E5%AF%BC%E8%BF%87%E7%A8%8B%E5%9C%A8%E7%99%BE%E5%BA%A6%E7%99%BE%E7%A7%91%E4%B8%8A%E7%9C%8B%E8%A7%81%EF%BC%9A%E2%80%9C%E7%94%B1%E4%B8%80%E5%85%83%E4%B8%89%E6%AC%A1%E6%96%B9%E7%A8%8B%E7%9A%84%E5%AE%8C%E6%95%B4%E5%BC%8FX3%2Ba1X2%2Ba2X%2Ba3%3D0+%281%29%E5%92%8C%E7%BC%BA%E9%A1%B9%E5%BC%8FX3%2BpX%2Bq%3D0+%282%29%E5%8F%AF%E7%9F%A5%2C%E6%AC%B2%E5%B0%86%E5%BC%8F+%281%29%E8%BD%AC%E6%8D%A2%E4%B8%BA%E5%BC%8F+%282%29%2C%E9%9C%80%E4%BB%A4y%3DX%EF%BC%8Da1%2F3%E4%BB%A3%E5%85%A5%E5%BC%8F+%281%29%2C%E5%BE%97%EF%BC%88X%EF%BC%8Da1%2F3%EF%BC%893%2Ba1%EF%BC%88X%EF%BC%8Da1%2F3%EF%BC%892%2B%E2%80%A6%3D0%2C%E2%80%9D%E5%85%B6)
卡当公式推导过程在百度百科上看见:“由一元三次方程的完整式X3+a1X2+a2X+a3=0 (1)和缺项式X3+pX+q=0 (2)可知,欲将式 (1)转换为式 (2),需令y=X-a1/3代入式 (1),得(X-a1/3)3+a1(X-a1/3)2+…=0,”其
卡当公式推导过程
在百度百科上看见:
“
由一元三次方程的完整式X3+a1X2+a2X+a3=0 (1)
和缺项式X3+pX+q=0 (2)可知,
欲将式 (1)转换为式 (2),
需令y=X-a1/3代入式 (1),
得(X-a1/3)3+a1(X-a1/3)2+…=0,
”
其中y=X-a1/3 是什么,怎么来的,是a1除以3吗?
卡当公式推导过程在百度百科上看见:“由一元三次方程的完整式X3+a1X2+a2X+a3=0 (1)和缺项式X3+pX+q=0 (2)可知,欲将式 (1)转换为式 (2),需令y=X-a1/3代入式 (1),得(X-a1/3)3+a1(X-a1/3)2+…=0,”其
是的,作用是消除二次项
x^3 + a1*x^2 + a2*x + a3 = 0…………(1)
设y=x+a1/3,则x=y-a1/3,代入(1)式,得:
(y-a1/3)^3 + a1*(y-a1/3)^2 + a2*(y-a1/3) + a3 = 0
y^3 - 3y^2*a1/3 + 3y*a1^2/9 - a1^3/27 + a1(y^2 - 2y*a1/3 + a1^2/9) + a2*(y-a1/3) + a3 = 0
y^3 - a1*y^2 + a1^2*y/3 - a1^3/27 + a1*y^2 - 2y*a1^2/3 + a1^3/9 + a2*y - a2*a1/3 + a3 = 0
y^3 + (-a1+a1)y^2 + (a1^2/3-2a1^2/3+a2)y + (-a1^3/27+a1^3/9-a2*a1/3+a3) = 0
y^3 + (a2-a1^2/3)y + (2a1^3/27-a1*a2/3+a3) = 0…………(2)
令p=a2-a1^2/3,q=2a1^3/27-a1*a2/3+a3,则(2)式可化为
y^3+py+q=0