（1）a+b+

2b2

a−b

=

(a+b)(a−b)

a−b

+

2b2

a−b

=

a2+b2

a−b

；
（2）(x+1−

3

x−1

)÷

x+2

2x−2

=[

(x+1)(x−1)

x−1

-

3

x−1

]×

2(x−1)

x+2

=

(x+2)(x−2)

x−1

×

2(x−1)

x+2

=2x-4；
（3）(2+

1

a−1

−

1

a+1

)÷(a−

a

1−a2

)，
=[

2(a+1)(a−1)

(a+1)(a−1)

+

(a+1)

(a−1)(a+1)

-

a−1

(a+1)(a−1)

]÷[

a(a+1)(a−1)

(a+1)(a−1)

+

a

(a+1)(a−1)

]
=

2a2

(a+1)(a−1)

×

(a+1)(a−1)

a3

=

2

a

，
把a=2代入原式得：原式=

2

a

=1．