(1)原式=
x−2
x+3]•
(x+3)(x−3)
(x+2)(x−2)
=[x−3/x+2];
(2)原式=[3/x−4]-[24
(x+4)(x−4)
=
3(x+4)−24
(x+4)(x−4)
=
3x−12
(x+4)(x−4)
=
3/x+4];
(3)原式=[1/2x]-[1/x+y]•
(x+y)(1−2x)
2x
=[1/2x]-[1−2x/2x]
=[1−1+2x/2x]
=1.
(1)原式=
x−2
x+3]•
(x+3)(x−3)
(x+2)(x−2)
=[x−3/x+2];
(2)原式=[3/x−4]-[24
(x+4)(x−4)
=
3(x+4)−24
(x+4)(x−4)
=
3x−12
(x+4)(x−4)
=
3/x+4];
(3)原式=[1/2x]-[1/x+y]•
(x+y)(1−2x)
2x
=[1/2x]-[1−2x/2x]
=[1−1+2x/2x]
=1.