数学题|x-(a+1)^2⼀2|<=(a-1)^2⼀2的每一个解都使x^2-3(a+1)x+2(3a+1)<=0成立，求a范围

解：x^2-3(a+1)x+2(3a+1)=(x-2)(x-(3a+1))<=0, -(a-1)^2/2<=x-(a+1)^2/2<=(a-1)^2/2.
2a<=x<=a^2+1,则a<= -1或1<=a<=3

1 -(a-1)^2/2 <= x-(a+1)^2/2 <= (a-1)^2/2
2a = -(a-1)^2/2 +(a+1)^2/2 <= x <= (a-1)^2/2+(a+1)^2/2= a^2 +1

2a <= x <= a^2 +1

2 （x－1.5(a+1) )^2 <= 长度受限，自己解2吧

一式得2a≤x≤a^2+1
二式得(x-2)(x-3a-1)≤0
(1)a＜1/3,则解为3a+1≤x≤2
得3a+1≤2a 且a^2+1≤2
得a=-1
(2)a=1/3不成立
(3)a＞1/3则解为2≤x≤3a+1
2≤2a且a^2+1≤3a+1
得1≤a≤3
所以a=-1或1≤a≤3

前面不等式的解集是后面不等式的子集
由于（a-1)^2/2>=0，前面的解集是
-(a-1)^2/2= 2a=后面的
x^2-3(a+1)x+2(3a+1)=[x-(3a+1)](x-2)<=0
然后讨论3a+1与2的关系，求解

由已知得：2a≤x≤a^2＋1
又(x－2)[x－(3a＋1)]≤0 ⑴
两种情况：2≤2a且3a＋1≥a^2＋1，2≤3a＋1，解得：1≤a≤3
或3a＋1≤2a且a^2＋1≤2，2≥3a＋1，解得：a＝－1
故a＝－1或1≤a≤3

|x-(a+1)^2/2|<=(a-1)^2/2的解2a <= x <= a^2+1
x1=3a+1, x2=2
(1)a^2+1 <=2 ,-1<=a<=1; 3a+1<=2a , a<=-1; a=-1
(2) a^2+1<=3a+1,0<=a<=3;2a>=2 , a>=1;1<=a<=3
所以: 1<=a<=3且a=-1