A=｛x｜x^2+ax+1<=0｝,
B={x｜x^2-3x+2<=0}={x|1<=x<=2}
A屬于B
則不等式x^2+ax+1<=0的解應(yīng)在1<=x<=2的范圍內(nèi)
函數(shù)f(x)=x^2+ax+1的圖像開口向上,根據(jù)題意,有：
當(dāng)1<=-a<=2,即-2<=a<=-1時(shí),
f(1)>=0
f(2)>=0
a+2>=0
2a+5>=0
解不等式組,得a>=-2,
綜合可得-2<=a<=-1
當(dāng)a<-2或a>-1時(shí),
判別式<=0
a²-4<=0
-2<=a<=2
綜合可得-1綜上所述,實(shí)數(shù)a的取值范圍是：-2<=a<=2