一元二次方程解答题不解方程,作一个一元二次方程,使它的两根:(1)分别是x²－2x－3＝0的两根立方（2）分别是方程3x

x^2-2x-3=0
x1+x2=2
x1x2=-3
x1`=x1^3
x2`=x2^3
x1`+x2`=x1^3+x2^3=(x1+x2)(x1^2-x1x2+x2^2)=(x1+x2)((x1+x2)^2-2x1x2-x1x2)
=(x1+x2)((x1+x2)^2-3x1x2)
=2(2^2-3*(-3))=2(4+9)=2*13=26
x1`x2`=x1^3*x2^3=(x1x2)^3=(-3)^3=-27
x^2-26x-27=0
(2)3x^2+5x-10=0
x1+x2=-5/3
x1x2=-10/3
x1`=3x1,x2`=3x2
x1`+x2`=3x1+3x2=3(x1+x2)=3*(-5/3)=-5
x1`x2`=3x1*3x2=9x1x2=9*(-10/3)=-30
x^2+5x-30=0
(3)5x^2+2x-3=0
x1+x2=-2/5
x1x2=-3/5
x1`=-1/x1^2,x2`=-1/x2^2
x1`+x2`=-1/x1^2-1/x2^2=-(1/x1^2+1/x2^2)=-(x1^2+x2^2)/x1^2x2^2
=-((x1+x2)^2-2x1x2)/(x1x2)^2
=-(4/25-2*(-3/5))/9/25
=-(4/25+6/5)/9/25=-(34/25)/9/25=-34/9
x1`x2`=-1/x1^2*(-1/x2^2)=1/x1^2x2^2=1/(x1x2)^2=1/(-3/5)^2=1/9/25=25/9
x^2+34/9x+25/9=0
9x^2+34x+25=0

利用韦达定理
（1）设已知方程的两个根分别是x1和x2
则x1+x2=2 x1x2=-3
所求的方程的两个根分别是x1*3+x2*3
=(x1+x2)(x1*2-x1x2+x2*2)
...

（1）因为x1+x2=2,x1*x2=-3
所以x1^3+x2^3=(x1+x2)(x1^2-x1*x2+x2^2)=-3*[(x1+x2)^2-3x1*x2]
=-3*[4-3*(-3)]
=-39
x1^3*x2^3=(x1*x2)^3=(-3)^3=-27
所以方程...