试题

题目：
化简：
（1）
x3+xy2+1
x3+xy-x2y-y3
·
x2y-xy2
x3+xy2-x2y-y3
+
x2y+xy2
x3+x2y+xy2+y3
·
x2y+y3+1
x2y+y3-x3-xy2
+
x3+y3
x3+x2y+xy2+y3

（2）
1
x(x+a)
+
1
(x+a)(x+2a)
+
1
(x+2a)(x+3a)
+
1
(x+3a)(x+4a)

（3）(
b2-bc+c2
a
+
a2
b+c
-
3
1
b
+
1
c
)×
2
b
+
2
c
1
bc
+
1
ca
+
1
ab
+(a+b+c)2
（4）
1
(x-1)(x-2)
+
1
(x-2)(x-3)
+…+
1
(x-n)(x-n-1)
．
答案
解：（1）
原式=
x3+xy2+1
x2(x-y)+y2(x-y)
·
xy(x-y)
x2(x-y)+y2(x-y)
+
xy(x+y)
x2(x+y)+y2(x+y)
·
x2y+y3+1
y2(y-x)+x2(y-x)
+
(x+y)(x2-xy+y2)
x2(x+y)+y(x+y)
=
xy(x3+xy2+1)
(x2+y2)(x-y)(x2+y2)
+
xy(x2y+y3+1)
(x2+y2)(x2+y2)(y-x)
+
x2-xy+y2
x2+y2
=
xy[x2(x-y)+y2(x-y)]
(x-y)(x2+y2)2
+
x2-xy+y2
x2+y2
=
x2+y2
x2+y2
=1

（2）∵
a
x(x+a)
=
1
x
-
1
x+a
，
a
(x+a)(x+2a)
=
1
x+a
-
1
x+2a
，
a
(x+2a)(x+3a)
=
1
x+a
-
1
x+2a
，
a
(x+3a)(x+4a)
=
1
x+3a
-
1
x+4a
，

∴原式=
1
a
(
1
x
-
1
x+a
)+
1
a
(
1
x+a
-
1
x+2a
)+
1
a
(
1
x+2a
-
1
x+3a
)+
1
a
(
1
x+3a
-
1
x+4a
)
=
1
a
(
1
x
-
1
x+4a
)

=
4
x(x+4a)

（3）原式=(
b2-bc+c2
a
+
a2-3bc
b+c
)
2(b+c)
bc
a+b+c
abc
+(a+b+c)2
=(
b2-bc+c2
a
+
a2-3bc
b+c
)·
2(b+c)a
a+b+c
+(a+b+c)2
=
2(b+c)(b2-bc+c2)
a+b+c
+
2a(a2-3bc)
a+b+c
+(a+b+c)2
=
2b3+2c3+2a3-6abc
a+b+c
+(a+b+c)2
=
2(a3+b3+c3-3abc)+(a+b+c)3
a+b+c
=
2(a+b+c)(a2+b2+c2-ab-ac-bc)+(a+b+c)3
a+b+c
=
(a+b+c)(2a2+2b2+2c2-2ab-2ac-2bc+a2+b2+c2+2ab+2ac+2bc)
a+b+c
=
(a+b+c)(3a2+3b2+3c2)
a+b+c
=3a2+3b2+3c2

（4）∵
1
(x-1)(x-2)
=
1
x-2
-
1
x-1
1
(x-2)(x-3)
=
1
x-3
-
1
x-2
…
1
(x-n)[x-(n+1)]
=
1
x-(n+1)
-
1
x-n

∴原式=
1
x-2
-
1
x-1
+
1
x-3
-
1
x-2
+…+
1
x-(n+1)
-
1
x-n

=
1
x-(n+1)
-
1
x-1
=
n
(x-1)[x-(n+1)]

解：（1）
原式=
x3+xy2+1
x2(x-y)+y2(x-y)
·
xy(x-y)
x2(x-y)+y2(x-y)
+
xy(x+y)
x2(x+y)+y2(x+y)
·
x2y+y3+1
y2(y-x)+x2(y-x)
+
(x+y)(x2-xy+y2)
x2(x+y)+y(x+y)
=
xy(x3+xy2+1)
(x2+y2)(x-y)(x2+y2)
+
xy(x2y+y3+1)
(x2+y2)(x2+y2)(y-x)
+
x2-xy+y2
x2+y2
=
xy[x2(x-y)+y2(x-y)]
(x-y)(x2+y2)2
+
x2-xy+y2
x2+y2
=
x2+y2
x2+y2
=1

（2）∵
a
x(x+a)
=
1
x
-
1
x+a
，
a
(x+a)(x+2a)
=
1
x+a
-
1
x+2a
，
a
(x+2a)(x+3a)
=
1
x+a
-
1
x+2a
，
a
(x+3a)(x+4a)
=
1
x+3a
-
1
x+4a
，

∴原式=
1
a
(
1
x
-
1
x+a
)+
1
a
(
1
x+a
-
1
x+2a
)+
1
a
(
1
x+2a
-
1
x+3a
)+
1
a
(
1
x+3a
-
1
x+4a
)
=
1
a
(
1
x
-
1
x+4a
)

=
4
x(x+4a)

（3）原式=(
b2-bc+c2
a
+
a2-3bc
b+c
)
2(b+c)
bc
a+b+c
abc
+(a+b+c)2
=(
b2-bc+c2
a
+
a2-3bc
b+c
)·
2(b+c)a
a+b+c
+(a+b+c)2
=
2(b+c)(b2-bc+c2)
a+b+c
+
2a(a2-3bc)
a+b+c
+(a+b+c)2
=
2b3+2c3+2a3-6abc
a+b+c
+(a+b+c)2
=
2(a3+b3+c3-3abc)+(a+b+c)3
a+b+c
=
2(a+b+c)(a2+b2+c2-ab-ac-bc)+(a+b+c)3
a+b+c
=
(a+b+c)(2a2+2b2+2c2-2ab-2ac-2bc+a2+b2+c2+2ab+2ac+2bc)
a+b+c
=
(a+b+c)(3a2+3b2+3c2)
a+b+c
=3a2+3b2+3c2

（4）∵
1
(x-1)(x-2)
=
1
x-2
-
1
x-1
1
(x-2)(x-3)
=
1
x-3
-
1
x-2
…
1
(x-n)[x-(n+1)]
=
1
x-(n+1)
-
1
x-n

∴原式=
1
x-2
-
1
x-1
+
1
x-3
-
1
x-2
+…+
1
x-(n+1)
-
1
x-n

=
1
x-(n+1)
-
1
x-1
=
n
(x-1)[x-(n+1)]
考点梳理
分式的混合运算．
（1）先对分子、分母进行因式分解，再约分，最后通分，实现化简．
（2）观察发现
a
x(x+a)
=
1
x
-
1
x+a
，
a
(x+a)(x+2a)
=
1
x+a
-
1
x+2a
，
a
(x+2a)(x+3a)
=
1
x+a
-
1
x+2a
，
a
(x+3a)(x+4a)
=
1
x+3a
-
1
x+4a
，
因而将原式转化为
1
a
(
1
x
-
1
x+a
)+
1
a
(
1
x+a
-
1
x+2a
)+
1
a
(
1
x+2a
-
1
x+3a
)+
1
a
(
1
x+3a
-
1
x+4a
)，再通过提取公因式，求解；
（3）首先将-
3
1
b
+
1
c
转化为-
bc
b+c
再与
a2
b+c
通分，将
2
b
+
2
c
1
bc
+
1
ca
+
1
ab
转化为
2(b+c)a
a+b+c
，再利用乘法的分配律，进一步通过通分，最终达到目的；
（4）观察发现
1
(x-1)(x-2)
=
1
x-2
-
1
x-1
，
1
(x-2)(x-3)
=
1
x-3
-
1
x-2
…
1
(x-n)[x-(n+1)]
=
1
x-(n+1)
-
1
x-n
1
(x-2)(x-3)
=
1
x-3
-
1
x-2
，…，
1
(x-n)[x-(n+1)]
=
1
x-(n+1)
-
1
x-n

将这些式子代入原式．加减抵消相同项，最终得到结果．
本题考查分式的混合运算，关键是通分，合并同类项，注意混合运算的运算顺序．同学们特别注意（2）中的
a
x(x+a)
=
1
x
-
1
x+a
，
a
(x+a)(x+2a)
=
1
x+a
-
1
x+2a
，
a
(x+2a)(x+3a)
=
1
x+a
-
1
x+2a
，
a
(x+3a)(x+4a)
=
1
x+3a
-
1
x+4a
，
（4）中
1
(x-1)(x-2)
=
1
x-2
-
1
x-1
，
1
(x-2)(x-3)
=
1
x-3
-
1
x-2
…
1
(x-n)[x-(n+1)]
=
1
x-(n+1)
-
1
x-n
1
(x-2)(x-3)
=
1
x-3
-
1
x-2
，…，
1
(x-n)[x-(n+1)]
=
1
x-(n+1)
-
1
x-n

有效转化．
计算题．
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