计算. (1)a+b+2b2a−b; (2)(x+1−3/x−1)÷x+2/2x−2; (3)(2+1/a−1−1/a+1)÷(a−a/1−a2),其中a=2.

问题描述:

计算.
(1)a+b+

2b2
a−b

(2)(x+1−
3
x−1
x+2
2x−2

(3)(2+
1
a−1
1
a+1
)÷(a−
a
1−a2
)
,其中a=2.

(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.