f(x)=(3x+a)/(x+2)= (3x+6+a-6)/(x+2)
=3+(a-6)/(x+2)
显然,函数1/(x+2) 在区间(-2,+∞)上单调递减,
若使函数f(x)=在区间(-2,+∞)上单调递减,
则需a-6>0,即a>6.