[9x+(1除以3√x)]^12 的展开式中,常数项之系数为?

那是3次根号下x吗?
[9x+(1/³√x)]^12
通项
Tr+1=C(12,r)(9x)^(12-r)(1/³√x)^r
=C(12,r)*9^(12-r) x^(12-r)*x^(-r/3)
=9^(12-r)C(12,r)x^(12-4r/3)
令12-4r/3=0得r=9
∴常数项为T10=9^3C(12,9)
=729*12*11*10/(3*2)
=16038
[9x+(1/3√x)]^12
Tr+1=C(12,r)(9x)^(12-r)(1/3√x)^r
=C(12,r)*9^(12-r)（1/3)^r* x^(12-r)*x^(-r/2)
=9^(12-r)*1/3^r C(12,r)x^(12-3r/2)
令12-3r/2=0得r=8
∴常数项为T9=9^4*1/3^8*C(12,8)
=12*11*10*9/(4*3*2*1)
=495