Question

計算：
（1）

a

a3+a2b+ab2+b3

+

b

a3-a2b+ab2-b3

+

1

a2-b2

-

1

a2+b2

-

a2+3b2

a4-b4

．
（2）

x3-1

x3+2x2+2x+1

+

x3+1

x3-2x2+2x-1

-

2(x2+1)

x2-1

（3）

b2 a2 + a2 b2 +2

b3 a3 - a3 b3 -3( b a - a b )

÷

b a + a b

b2 a2 + a2 b2 -2

．

Answer 1

考點：分式的混合運算

專題：

分析：首先將所給的分式分子、分母因式分解，然后化簡求值，即可解決問題．

解答：解：（1）原式=

a

a2(a+b)+b2(a+b)

+

b

a2(a-b)+b2(a-b)

+

a2+b2

a4-b4

-

a2-b2

a4-b4

-

a2+3b2

a4-b4

=

a(a-b)+b(a+b)

(a+b)(a2+b2)(a-b)

+

-(a2+b2)

(a2+b2)(a2-b2)

=

a2+b2

(a2+b2)(a2-b2)

-

1

a2-b2

=

1

a2-b2

-

1

a2-b2

=0．
（2）原式=

x3-1

x3+2x2+2x+1

+

x3+1

x3-2x2+2x-1

-

2(x2+1)

x2-1

=

x3-1

x2(x+1)+(x+1)2

+

x3+1

x2(x-1)-(x-1)2

-

2(x2+1)

x2-1

=

(x-1)(x2+x+1)

(x+1)(x2+x+1)

+

(x+1)(x2-x+1)

(x-1)(x2-x+1)

-

2x2+2

x2-1

=

x-1

x+1

+

x+1

x-1

-

2x2+2

x2-1

=

(x-1)2

x2-1

+

(x+1)2

x2-1

-

2x2+2

x2-1

=

x2-2x+1+x2+2x+1-2x2-2

x2-1

=0．
（3）原式=

b2 a2 + a2 b2 +2

b3 a3 - a3 b3 -3( b a - a b )

÷

b a + a b

b2 a2 + a2 b2 -2

=

( b a + a b )2

( b a - a b )( b2 a2 + a2 b2 +1)-3( b a - a b )

÷

b a + a b

( b a - a b )2

=

( b a + a b )2

( b a - a b )( b a - a b )2

×

( b a - a b )2

b a + a b

=

b a + a b

b a - a b

=

a2+b2

b2-a2

．

點評：該題主要考查了分式的混合運算法則及其應用問題；解題的關鍵是準確運用因式分解法將所給的分式分子、分母因式分解，然后化簡求值．