计算下列各式:(1)1a-b+1a+b+2aa2+b2+4a3a4+b4;(2)x2+yzx2+(y-z)x-yz+y2-zxy2+(z+x)y+zx+z2+x

计算下列各式:(1)1a-b+1a+b+2aa2+b2+4a3a4+b4;(2)x2+yzx2+(y-z)x-yz+y2-zxy2+(z+x)y+zx+z2+x

题型:解答题难度:一般来源:不详
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.
举一反三
在公式
1
R1
+
1
R2
=
1
R
中,已知R1、R2,且R1+R2≠0,则R=______.
题型:填空题难度:一般| 查看答案
(1)观察下列各式:
1
6
=
1
2×3
=
1
2
-
1
3
1
12
=
1
3×4
=
1
3
-
1
4
1
20
=
1
4×5
=
1
4
-
1
5
1
30
=
1
5×6
=
1
5
-
1
6
,…
由此可推导出
1
42
=______.
(2)请猜想出能表示(1)的特点的一般规律,用含字母m的等式表示出来(m表示整数);
(3)请直接用(2)中的规律计算:
1
(x-2)(x-3)
-
2
(x-1)(x-3)
+
1
(x-1)(x-2)
的结果.
题型:解答题难度:一般| 查看答案
已知分式:A=
2
x2-1
B=
1
x+1
+
1
1-x
.(x≠±1).下面三个结论:①A,B相等,②A,B互为相反数,③A,B互为倒数,请问哪个正确?为什么?
题型:解答题难度:一般| 查看答案
对于正数x,规定f(x)=
x
1+x
,比如 f(3)=
3
1+3
,f(
1
2
)=
1
2
1+
1
2
=
1
3
,则计算f(1)+f(2)+f(3)+…+f(99)+f(100)+f(
1
100
)
+f(
1
99
)
+…+f(
1
3
)
+f(
1
2
)
+f(1)=______.
题型:填空题难度:一般| 查看答案
计算
1
x-1
-
x
x-1
的结果为(  )
A.1B.2C.-1D.-2
题型:单选题难度:一般| 查看答案
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