Question

計(jì)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)

．

Answer 1

（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）設(shè)x-y=a，y-z=b，z-x=c，則

(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．