A/(X+2)-B/(A-2)=4X/(X²-4)
[A(X-2)]/[(X+2)(X-2)]-[B(X+2)]/[(X+2)(X-2)]=4X/(X²-4)
(AX-2A)/(X²-4)-(BX+2B)/(X²-4)=4X/(X²-4)
[(A-B)X-2(A+B)]/(X²-4)=4X/(X²-4)
因这运算是恒等,即A-B=4,A+B=0
即A+B=0
A/(X+2)-B/(A-2)=4X/(X²-4)
[A(X-2)]/[(X+2)(X-2)]-[B(X+2)]/[(X+2)(X-2)]=4X/(X²-4)
(AX-2A)/(X²-4)-(BX+2B)/(X²-4)=4X/(X²-4)
[(A-B)X-2(A+B)]/(X²-4)=4X/(X²-4)
因这运算是恒等,即A-B=4,A+B=0
即A+B=0