2020中考数学一轮基础考点训练05 分式

7. (2019通辽)先化简,再求值.

??5-2x≥1,1x2+2x1

÷2+,请从不等组?的整数解中选择一个你喜欢的求值. 1-xx-2x+1x+2?x+3>0?

参考答案

基础达标训练

1. B 2. D 3. D 4. D

a2+12a2+1-2a2-1(a-1)(a+1)

5. A 【解析】-====a-1.

a+1a+1a+1a+1a+1

2aa+22a-(a+2)a-2

6. B 【解析】原式=-==

(a+2)(a-2)(a+2)(a-2)(a+2)(a-2)(a+2)(a-2)=1. a+27. 2

4(x+2)(x-2)x448. 【解析】原式=- =-=. 2-xx-2x-2x-22-x2

a-1(a+1)29. 2019 【解析】原式=×=a+1,当a=2018时,原式=2018+1=2019.

a+1a-1

10. 解:原式=

1m- m(m-1)m-1

2

1m=- m(m-1)m(m-1)1-m= m(m-1)=

(1+m)(1-m)

m(m-1)

2

1+m=-. m(x-2)x-211. 解:原式=÷ (x+2)(x-2)x(x+2) =

2

x-2x(x+2)

· x+2x-2

=x.

12. 解:原式=

2(a+1)(a-1)1·+ a-12(a-2)2-a =

a+11

- a-2a-2a. =

a-2

13. 解:原式=[-m-2

m2

(m+2)(m-2)

]

m-2

·

(m+2)(m-2)

2(m+2)

m2-(m2-4)m-2=·

m-22

4m-2· m-22

=2.

14. 解:圆圆的解答不正确,正确解答结果如下: 4x原式=-

(x+2)(x-2)2(x+2)

(x+2)(x-2)

x2-4

(x+2)(x-2)

4x-(2x+4)-(x-4)= (x+2)(x-2)=

-x(x-2)

(x+2)(x-2)

2

=-

xx+2

. a-1a-1

15. 解:原式= 2-

(a-1)(a+1)(a-1)

11- a-1a+1

a+1-(a-1)

(a-1)(a+1)

2. a-1

2

22

当a=3时,原式===1. 2

3-1(3)-116. 解:原式=

a(a+1)(a-1)a·-

a(a-1)a+1a-1

=1-aa-1

=-

1

. a-1

当a=2时, 1

原式=-=-1.

2-1

2x12(x+2)

17. 解:原式=[-]· (x+2)(x-2)x-23x=

2x-x-2

·

(x+2)(x-2)

2(x+2)

3x=

x-2

·

(x+2)(x-2)

2(x+2)

3x=2. 3x当x=-3时,

22

原式==-.

3×(-3)9能力提升拓展

(x+2)1(x+2)1111. B 【解析】化简2-==1-,∵x为正整数,即x≥1,∴0<2-x+4x+4x+1(x+2)x+1x+1x+1

2

2

11111≤,∴-≤-<0,∴≤1-<1,即原式的值落在段②之间. 22x+12x+1

x2+xy-xy+y2

2. B 【解析】根据去括号规则,第②步应为,故选B.

(x-y)(x+y)

3. ④

115m+5n-2mn5(m+n)-2mn10mn-2mn8mn4. -4 【解析】由+=2得m+n=2mn,∴====

mn-m-n-(m+n)-2mn-2mn-4.

a2-2ab+b2a2-ab2

5. 解:原式=÷- 22

a-baa+b(a-b)a2

=·- (a+b)(a-b)a(a-b)a+b=

12- a+ba+b1. a+b2

2

=-

∵(a-2)+b+1=0, ∴a-2=0,b+1=0. ∴a=2,b=-1.

1

当a=2,b=-1时,原式=-=-1.

2-1

(a+2)(a-2)1a(a-2)

6. 解:原式=[+]· 2

(a-2)a-22=(

a+21a(a-2)

+)· a-2a-22

a+3a(a-2)· a-22a2+3a2

. 2

∵a满足a+3a-2=0, ∴a+3a=2.

2

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