集合A={x|x²-[2a+(a²+1)]x+2a(a²+1)≤0},B=｛x|(x-2)[x-(3a+1)]≤0｝当实数a为何值时,A属于B

集合A={x|x²-[2a+(a²+1)]x+2a(a²+1)≤0},B=｛x|(x-2)[x-(3a+1)]≤0｝当实数a为何值时,A属于B
集合A={x|x²-[2a+(a²+1)]x+2a(a²+1)≤0},B=｛x|(x-2)[x-(3a+1)]≤0｝当实数a为何值时,A属于B

集合A={x|x²-[2a+(a²+1)]x+2a(a²+1)≤0},B=｛x|(x-2)[x-(3a+1)]≤0｝当实数a为何值时,A属于B
①
(x-2a)[x-(a²+1)]≤0;
∵a²+1≥2a;
∴ 2a≤x≤a²+1;
②当a>1/3;
(x-2)[x-(3a+1)]≤0;
2≤x≤3a+1;
∴2a≥2;
3a+1≥a²+1
∴1≤a≤3
③ 当a≤1/3;
(x-2)[x-(3a+1)]≤0;
3a+1≤x≤2;
∴2a≥3a+1;
2≥a²+1;
∴a=-1;
∴综上得：
a=-1或1≤a≤3