設(shè)m是不小于-1的實(shí)數(shù),使得關(guān)于x的方程x^+2(m-2)x+m^-3m+3=0有兩個(gè)不同的實(shí)數(shù)根,x1,x2,
1.若x1^+x2^=6,求m
2.求mx1^/(1-x1)+mx2^/(1-x2)的最大值
1:
x^+2(m-2)x+m^-3m+3=0
x1+x2=-2(m-2)
x1*x2=m^-3m+3
6=x1^+x2^=(x1+x2)^-2x1x2=2m^-10m+10
m^-5m+2=0 ==>m1=(5-17^0.5)/2 m2=(5+17^0.5)/2
因?yàn)?-1≤m≤1
所以:m1=(5-17^0.5)/2
2:
mx1^/(1-x1)+mx2^/(1-x2)=m(x1^-x1^x2+x2^-x1x2^)/(1-x1-x2+x1x2)
=m[(x1+x2)^-x1x2(x1+x2+2)]/[1-(x1+x2)+x1x2]
=m[(4-2m)^-(m^-3m+3)(4-2m+2)]/[1-(4-2m)+m^-3m+3]
=m(2m^3-8m^+8m-2)/[m(m-1)]
=2m(m-1)(m^-3m+1)/[m(m-1)]
=2[(m-3/2)^-9/4+1]
=2(m-3/2)^-5/2 ==>m=-1
最大值為:10