A=｛x|x2-ax+a2-19=0},
B={x|x2-5x+6=0}
= {x| (x-2)(x-3) =0 }
= {2,3}
C={x|x2+2x-8=0}
={ x| (x+4)(x-2)=0 }
= {-4,2}
if A∩B=ø
=> 2,3 不属于A
for x=2
4-2a+a2-19 ≠ 0
a2-2a-15≠ 0
(a-5)(a+3)≠ 0
a ≠ 5 or a ≠ -3
for x=3
9-3a+a2-19 ≠ 0
a2-3a-10 ≠ 0
(a-5)(a+2)≠ 0
a ≠ 5 and a ≠ -2
and A∩C=ø
=> -4,2 不属于 A
for x=-4
16+4a+a2-19 ≠ 0
a2+4a-3 ≠ 0
a ≠ -2+√7 or and a ≠ -2-√7
for x=2,we have a ≠ 5 or a ≠ -3
we have
(a ≠ 5 or a ≠ -3) and (a ≠ 5 and a ≠ -2) and (a ≠ -2+√7 or and a ≠ -2-√7)
ie a ≠ 5 and a ≠ -3 and a ≠ -2 and a ≠ -2+√7 and a ≠ -2-√7 #