令a=3^x
则a>0
y=(2a+1)/(4a-3)
=(2a-3/2+5/2)/(4a-3)
=(2a-3/2)/(4a-3)+(5/2)/(4a-3)
=1/2+5/(8a-6)
a>0
所以8a-6>-6
1/(8a-6)0
5/(8a-6)0
1/2+5/(8a-6)1/2
值域(-∞-1/3)∪(1/2,+∞)
令a=3^x
则a>0
y=(2a+1)/(4a-3)
=(2a-3/2+5/2)/(4a-3)
=(2a-3/2)/(4a-3)+(5/2)/(4a-3)
=1/2+5/(8a-6)
a>0
所以8a-6>-6
1/(8a-6)0
5/(8a-6)0
1/2+5/(8a-6)1/2
值域(-∞-1/3)∪(1/2,+∞)