问题 解答题
计算下列各式:
(1)
1
a-b
+
1
a+b
+
2a
a2+b2
+
4a3
a4+b4

(2)
x2+yz
x2+(y-z)x-yz
+
y2-zx
y2+(z+x)y+zx
+
z2+xy
z2-(x-y)z-xy

(3)
x3-1
x3+2x2+2x+1
+
x3+1
x3-2x2+2x-1
-
2(x2+1)
x2-1

(4)
(y-x)(z-x)
(x-2y+z)(x+y-2z)
+
(z-y)(x-y)
(x+y-2z)(y+z-2x)
+
(x-z)(y-z)
(y+z-2x)(x-2y+z)
答案

(1)

1
a-b
+
1
a+b
+
2a
a2+b2
+
4a3
a4+b4

=

2a
a2-b2
+
2a
a2+b2
+
4a3
a4+b4

=

4a3
a4-b4
+
4a3
a4+b4

=

8a7
a8-b8

(2)

x2+yz
x2+(y-z)x-yz
+
y2-zx
y2+(z+x)y+zx
+
z2+xy
z2-(x-y)z-xy

=

x(x-z)+z(x+y)
(x+y)(x-z)
+
y(x+y)-x(y+z)
(x+y)(y+z)
+
z(y+z)-y(z-x)
(z-x)(y+z)

=

x
x+y
+
z
x-z
+
y
y+z
-
x
x+y
-
z
x-z
-
y
y+z

=0;

(3)

x3-1
x3+2x2+2x+1
+
x3+1
x3-2x2+2x-1
-
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)
-
2(x2+1)
(x+1)(x-1)

=

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

=0;

(4)设x-y=a,y-z=b,z-x=c,则

(y-x)(z-x)
(x-2y+z)(x+y-2z)
+
(z-y)(x-y)
(x+y-2z)(y+z-2x)
+
(x-z)(y-z)
(y+z-2x)(x-2y+z)

=-

ac
(a-b)(b-c)
-
ab
(b-c)(c-a)
-
cb
(c-a)(c-b)

=-

ac(c-a)+ab(a-b)+bc(b-c)
(a-b)(b-c)(c-a)

=

(a-b)(b-c)(c-a)
(a-b)(b-c)(c-a)

=1.

单项选择题
单项选择题