Question

设m是不小于-1的实数，并使方程x^2+2(m-2)x+m^2-3m+3=0有两个不相等的实数根

（1）若X1^2+X2^2=6求m的值（2）求m(x1^2+x2^-x1^2x2-x1x2^2)/x1x2-x1-x2+1

Answer 1

若方程x^2+2(m-2)x+m^2-3m+3=0有两个不相等的实数根
方程式判别式Δ=4(m-2)²-4(m²-3m+3)＞0
m＜1
∴-1≤m＜1
(1)x1+x2=-2(m-2)=4-2m
x1*x2=m²-3m+3
X1²+X2²=6
(x1+x2)²-2x1*x2=6
(4-2m)²-2(m²-3m+3)=6
m=(5±√17)/2
∵ -1≤m＜1
∴ m=(5-√17)/2

(2)m(x1^2+x2^-x1^2x2-x1x2^2)/x1x2-x1-x2+1
=m[(x1+x2)²-2x1x2-x1x2(x1+x2)]/x1x2-(x1+x2)+1
=m[(x1+x2)²-x1x2(x1+x2+2)]/(x1x2)-(x1+x2)+1
=m(x1+x2)²/(x1x2)-m(x1+x2+2)-(x1+x2)+1
=m(4-2m)²/(m²-3m+3)-m(4-2m+2)-(4-2m)+1
=2m²-19+(32m-58)/(m²-3m+3)

Answer 2

方程x^2+2(m-2)x+m^2-3m+3=0有两个不相等的实数根
△=4(m-2)^2-4(m^2-3m+3)>0，即4m^2-16m+16-4m^2+12m-12=-4m+4>0
解得m<1, 已知m≥-1，∴-1≤m<1
1、若X1^2+X2^2=(x1+x2)^2-2x1x2=[-2(m-2)]^2-2(m^2-3m+3)=6，
即4m^2-16m+16-2m^2+6m-6=6, 即2m^2-10m+4=0, 即m^2-5m+2=0
解得m=[5-√(25-8)]/2=(5-√17)/2 (另一解(5+√17)/2)>1舍弃)
2、

Answer 3

X1+x2=-B/A X1*X2=C/A