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.

选择题

课内文言文知识积累(10分,每小题2分)

小题1:下列各组句子中不含通假字的一组是(   )

A.行李之往来,共其乏困邹忌修八尺有余

B.莫春者春服既成颁白者不负戴于道路矣

C.则无望民之多于邻国也赢粮而景从

D.序八州而朝同列旦日不可不蚤自来谢项王小题2:下列句中加点词不属于古今异义的一项是(   )

A.行李之往来,供其乏困

B.七十者可以食肉矣

C.而从六国破灭之故事

D.山东豪俊遂并起而亡秦族矣小题3:对下列句中加点的“其”意义和用法的分类,正确的一组是(   )

①问其深,则其好游者不能穷也   ②而余亦悔其之而不得极夫游之乐也   ③其闻道也亦先乎吾   ④夫庸知其年之先后生于吾乎   ⑤其皆出于此乎   ⑥其可怪也欤   ⑦其孰能讥之乎   ⑧既其出,则或咎其欲出者   ⑨其为惑也终不解矣

A.①②③④/⑤⑥⑦/⑧⑨

B.①⑧⑨/②/③④/⑤⑥⑦

C.①⑧⑨/②/③④/⑤/⑥⑦

D.①③④/②/⑤⑥⑦/⑧⑨小题4:下列句中加点的词活用方式不同的一项是(   )

A.群臣吏民,能面刺寡人之过者

B.后世之谬其传而莫能名者

C.项伯乃夜驰之沛公军

D.常以身翼蔽沛公小题5:对下列个句的句式分类正确的一组是(   )

①若亡郑而有益于君   ②夫晋何厌之有   ③申之以孝悌之义   ④沛公安在   ⑤沛公居山东时,贪于财物   ⑥凌万顷之茫然   ⑦渺渺兮予怀   ⑧洎牧以谗诛   ⑨为国者无使为积威之所劫哉

A.①③⑤/②④/⑥/⑦/⑧⑨

B.①②③/④⑥/⑤⑦/⑧⑨

C.①②③/④⑤/⑥⑧/⑦⑨

D.①③⑤/②⑥/④⑦/⑧⑨

单项选择题