a/(x+2)+b/(x-2)
=[a(x-2)+b(x+2)]/[(x+2)(x-2)]
=[(a+b)x+(2b-2a)]/(x²-4)
=4x/(x²-4)
所以(a+b)x+(2b-2a)=4x
对应项系数相等,常数项是0
所以a+b=4
2b-2a=0
a=2,b=2
a/(x+2)+b/(x-2)
=[a(x-2)+b(x+2)]/[(x+2)(x-2)]
=[(a+b)x+(2b-2a)]/(x²-4)
=4x/(x²-4)
所以(a+b)x+(2b-2a)=4x
对应项系数相等,常数项是0
所以a+b=4
2b-2a=0
a=2,b=2