利用根与系数关系求解

显示全部楼层 · 2010-8-22 19:40:58

若X1、X2是方程2X^2-3X+1=0的两个根，不解方程，求
1/X1^3+1/X2^3
X1^3-X2^3
关键是变化的过程，怎么变成韦达定理？

千问 · 2010-8-22 19:40:58

由韦达定理:x1+x2=3/2,x1x2=1/2,1/X1^3+1/X2^3=(x1^3+x2^3)/x1^3x2^3=(x1+x2)(x1^2-x1x2+x2^2)/(x1x2)^3=(x1+x2)[(x1+x2)^2-3x1x2]/(x1x2)^3=(3/2)(9/4-3/2)/(1/8)=9X1^3-X2^3=(x1-x2)(x1^2+x1x2+x2^2)=(x1-x2)[(x1+x2)^2-x1x2]=±√(x1-x2)^2[(9/4)-3/2]=±√[(x1+x2)^2-4x1x2](3/4)=±(1/2)*(3/4)=±3/8

千问 · 2010-8-22 19:40:58

因为X1、X2是方程2X^2-3X+1=0所以X1+X2=3/2，X1X2=1/2【韦达定理】所以：1/X1^3+1/X2^3=(X1^3+X2^3)/[(X1X2)^3]=(X1+X2)(X1^2-X1X2+X2^2)/[(X1X2)^3]=(X1+X2)[(X1+X2)^2-3X1X2]/[(X1X2)^3]=(3/2)[

千问 · 2010-8-22 19:40:58

x1+x2=3/2
x1x2=1/2(1)1/X1^3+1/X2^3=(x1^3+x2^3)/(x1x2)^3
=(x1+x2)(x1^2-x1x2+x2^2)/(x1x2)^3
=(x1+x2)[(x1+x2)^2-3x1x2)]/(x1x2)^3
带入即

千问 · 2010-8-22 19:40:58

x1+x2=3/2,x1*x2=1/2 x12+x22=5/41/X1^3+1/X2^3=(1/x1+1/x2)(1/x12-1/(x1*x2)+1/x22)1/X1^3-1/X2^3=(1/x1-1/x2)(1/x12+1/(x1*x2)+1/x22)带入解就可以了