解函數(shù)fx=x2-2x+1=（x-1）²對(duì)稱軸x=1
故a≥1時(shí),y=f（x）在區(qū)間[a,a+1]上是增函數(shù),即當(dāng)x=a時(shí),y有最小值f（a）=a²-2a+1
當(dāng)a+1≤1時(shí),即a≤0時(shí)y=f（x）在區(qū)間[a,a+1]上是減函數(shù),即當(dāng)x=a+1時(shí),y有最小值f（a+1）=a²
當(dāng)0＜a＜1時(shí),當(dāng)x=1時(shí),y有最小值f（1）=1²-2+1=0&#是啥意思你的手機(jī)字庫(kù)不全，顯示的與電腦不一致，請(qǐng)用電腦看。。。嗯