高一数学再线等解答

(1)任取x1,x2属于[1,正无穷),且x1<x2
则f(x1)-f(x2)
=(x1^2+ax1+a)/x1-[(x2^2+ax2+a)/x2]
=[x2*(x1^2+ax1+a)-x1*(x2^2+ax2+a)]/x1x2
=[x1^2*x2-x1*x2^2+ax2-ax1]/x1x2
=[x1x2(x1-x2)+a(x2-x1)]/x1x2
=[(x1-x2)(x1x2-a)]/x1x2
由1=<x1<x2,a<1
则x1-x2<0,x1x2-a>0,x1x2>0
则f(x1)-f(x2)<0
即1=<x1<x2时,f(x1)<f(x2)
则f(x)在[1,正无穷)上单调递增
(2)由f(3m)>f(5-2m),f(x)在[1,+无穷)上单调递增
则有: 3m>5-2m>=1
则解得:1<m<=2
(3)g(x)=x*f(x)
=x^2+ax+a
又g(x)+2x+3/2>0
则x属于[2,5]时,x^2+(a+2)x+(a+3/2)>0恒成立
则:t(x)=x^2+(a+2)x+(a+3/2)
t(2)>0,t(5)>0
则a>-19/6
[1]:判别式=<0
(a+2)^2-4(a+3/2)=<0
a^2+4a+4-4a-6=<0
a^2=<2
-根号2=<a<=根号2
[2]:判别式>0,
a>根号2 或 a<-根号2
又a>-19/6，则-19/6<a<-根号2或a>根号2

综合以上,则-19/6<a<=根号2