A=｛x｜x^2+ax+1<=0｝,
B={x｜x^2-3x+2<=0}={x|1<=x<=2}
A属于B
则不等式x^2+ax+1<=0的解应在1<=x<=2的范围内
函数f(x)=x^2+ax+1的图像开口向上,根据题意,有：
当1<=-a<=2,即-2<=a<=-1时,
f(1)>=0
f(2)>=0
a+2>=0
2a+5>=0
解不等式组,得a>=-2,
综合可得-2<=a<=-1
当a<-2或a>-1时,
判别式<=0
a²-4<=0
-2<=a<=2
综合可得-1综上所述,实数a的取值范围是：-2<=a<=2