设a,b,c是方程x^3+px+q=0 的三个根,行列式 a b c
设a,b,c是方程x^3+px+q=0 的三个根,行列式 a b c
c a b 等于多少?(要过程)
b c a
行列式 0 0 ... 0 1 0
0 0 ... 2 0 0
... ... ... ... ... ... 为什么是-1998!
0 1997 ... 0 0 0
1998 0 ... 0 0 0
0 0 ... 0 0 1
c a b 等于多少?(要过程)
b c a
行列式 0 0 ... 0 1 0
0 0 ... 2 0 0
... ... ... ... ... ... 为什么是-1998!
0 1997 ... 0 0 0
1998 0 ... 0 0 0
0 0 ... 0 0 1
赞 共回答了14个问题采纳率：92.9% 举报
x1=1/6*(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3)-2*p/(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3)
x2=-1/12*(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3)+p/(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3)+1/2*i*3^(1/2)*(1/6*(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3)+2*p/(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3))
x3=-1/12*(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3)+p/(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3)-1/2*i*3^(1/2)*(1/6*(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3)+2*p/(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3))
x2=-1/12*(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3)+p/(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3)+1/2*i*3^(1/2)*(1/6*(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3)+2*p/(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3))
x3=-1/12*(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3)+p/(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3)-1/2*i*3^(1/2)*(1/6*(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3)+2*p/(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3))
1年前
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