Question

求函数y=2x+1/1-3x的值域

Answer 1

y=(2x-2/3+5/3)/[-(3x-1)]
=(2x-2/3)[-(3x-1)]+(5/3)[-(3x-1)]
=2(x-1/3)/[-3(x-1/3)]+5/(3-9x)
=-2/3+5/(3-9x)
因为5/(3-9x)≠0
所以-2/3+5/(3-9x)≠5/3
所以值域(-∞，5/3)∪(5/3，+∞)

Answer 2

解y=(2x+1)/(1-3x)
=-(2x+1)/(3x-1)
=-[2/3(3x-1)+5/3]/(3x-1)
=-2/3+(-5/3)/(3x-1)
∵(-5/3)/(3x-1)≠0
故-2/3+(-5/3)/(3x-1)≠-2/3
故y≠-2/3
故函数的值域为{y/y≠-2/3}