函数f(x)=x+2分之ax+1(a属于R)在（-2,2）内为增函数,则实数a的范围是……

f(x)=(ax+1)/(x+2)
不妨设x1＞x2＞-2
因为f(x)在(-2,+2)上为增函数
则,f(x1)-f(x2)=(ax1+1)/(x1+2)-(ax2+1)/(x2+2)
=[(ax1+1)(x2+2)-(ax2+1)(x1+2)]/[(x1+2)(x2+2)]
=[(ax1x2+2ax1+x2+2)-(ax1x2+x1+2ax2+2)]/[(x1+2)(x2+2)]
=[(2a-1)(x1-x2)]/[(x1+2)(x2+2)]
＞0
上式中,x1-x2＞0,(x1+2)(x2+2)＞0
所以,2a-1＞0
所以,a＞1/2