设lgx>2/lgx+1.则x的取值范围是?

问题描述:

设lgx>2/lgx+1.则x的取值范围是?

lgx +1>0时,
lg(10x)>0
10x>1
x>1/10
lgx(lgx+1)>2
(lgx)²+lgx-2>0
(lgx+2)(lgx-1)>0
lgx>1或lgxx>10或x1/10,因此x>10

lgx +1lg(10x)00lgx(lgx+1)(lgx)²+lgx-2(lgx+2)(lgx-1)-21/100
综上,得1/10010

首先,x>0且x不等于1,;
然后,设m=lgx,解方程m>2/m+1.分情况讨论,m>0时,得m>2;m因而,解得x>100或0.1


lgx>2/(lgx+1)
令lgx=t
t>2/(t+1)
t-2/(t+1)>0
(t²+t-2)/(t+1)>0
(t+2)(t-1)/(t+1)>0
利用穿针引线法
-21
即-21
∴1/10010