设Χ1和Χ2是方程aΧ²+bΧ+c=0(a≠0) 求代数式:a(Χ1³+Χ2³)+b(Χ1²+Χ2²)+c(Χ1+Χ2)的值
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![设Χ1和Χ2是方程aΧ²+bΧ+c=0(a≠0) 求代数式:a(Χ1³+Χ2³)+b(Χ1²+Χ2²)+c(Χ1+Χ2)的值](/uploads/image/z/5541799-31-9.jpg?t=%E8%AE%BE%CE%A71%E5%92%8C%CE%A72%E6%98%AF%E6%96%B9%E7%A8%8Ba%CE%A7%26%23178%3B%2Bb%CE%A7%2Bc%3D0%EF%BC%88a%E2%89%A00%EF%BC%89+%E6%B1%82%E4%BB%A3%E6%95%B0%E5%BC%8F%EF%BC%9Aa%EF%BC%88%CE%A71%26%23179%3B%2B%CE%A72%26%23179%3B%EF%BC%89%2Bb%EF%BC%88%CE%A71%26%23178%3B%2B%CE%A72%26%23178%3B%EF%BC%89%2Bc%EF%BC%88%CE%A71%2B%CE%A72%EF%BC%89%E7%9A%84%E5%80%BC)
设Χ1和Χ2是方程aΧ²+bΧ+c=0(a≠0) 求代数式:a(Χ1³+Χ2³)+b(Χ1²+Χ2²)+c(Χ1+Χ2)的值
设Χ1和Χ2是方程aΧ²+bΧ+c=0(a≠0)
求代数式:
a(Χ1³+Χ2³)+b(Χ1²+Χ2²)+c(Χ1+Χ2)的值
设Χ1和Χ2是方程aΧ²+bΧ+c=0(a≠0) 求代数式:a(Χ1³+Χ2³)+b(Χ1²+Χ2²)+c(Χ1+Χ2)的值
Χ1和Χ2是方程aΧ²+bΧ+c=0(a≠0)的解
所以X1+X2=-b/a,X1*X2=c/a
Χ1²+Χ2²=(X1+X2)^2-2X1X2=b^2/a^2+2bc/a^2=(b^2+2bc)/a^2=(b+2c)b^2/a^2
Χ1³+Χ2³=(X1+X2)(X1^2-X1X2+X2^2)=-b/a*[(b^2+2bc)/a^2-c/a]=bc/a^2-(b+2c)b^2/a^3
Χ1+Χ2=-b/a
a(Χ1³+Χ2³)+b(Χ1²+Χ2²)+c(Χ1+Χ2)=bc/a-(b+2c)b^2/a^2+(b+2c)b^2/a^2-b/a=b(c-1)/a
原始=x1(ax1^2+bx1+c)+x2(ax2^2+bx2+c)=0
52
3333324234243152354342424242343才怪
Χ1³+Χ2³=(X1+X2)*(Χ1²+Χ2²-X1X2)=(X1+X2)*[(X1+x2)^2-3X1X2]
Χ1²+Χ2²=(X1+x2)^2-2X1X2
将Χ1+Χ2=-b/a X1X2=c/a代入代数式得
原式=c/a(b+c)
谢谢