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中考一元一次不等式组专题复习

中考一元一次不等式组专题复习
中考一元一次不等式组专题复习

九年级一元一次不等式(组)复习

一、知识点回顾及典型例题 (一)、不等式的定义: (二)、不等式的基本性质:

1.(2012广州市)已知a >b,c 为任意实数,则下列不等式中总是成立的是( ) A. a+c <b+c B. a -c >b -c C. ac <bc D. ac >bc

2.①若3<x ,则x 3; ②若-2<x ,则0 x +2; ③若-2a ≥8,则a 4; ④若x >y ,则m 2 x m 2 y 。

(三)、不等式的解和不等式的解集的定义:

⑴能使不等式成立的未知数的值(一个或几个),叫做不等式的解。 ⑵一个含有未知数的不等式的所有解,组成这个不等式的解集。 1.(2014衢州)不等式2x -1>1

2

x 的解集是 .

2:不等式53-x <x +3的正整数解有( ) 3. (2014攀枝花)下列说法中,错误..

的是( ) A. 不等式2-x 的解集是3->x D. 不等式10

(四)、一元一次不等式的定义和解法:

⑴不等式的左右两边都是整式,只含有一个未知数,并且未知数的最高次数是1,这样的不等式叫一元一次不等式。其标准形式:ax+b <0或ax+b ≤0,ax+b >0或ax+b ≥0(a ≠0).

⑵解一元一次不等式的一般步骤: 例:13

132

1≤---x x 解不等式: 4

12

33523+>--

x x ; 3252132x x x -≤--

(五)、一元一次不等式组:

⑴关于同一个未知数的几个一元一次不等式合在一起就组成一个一元一次不等式组。 ⑵一元一次不等式组中各个不等式的解集的公共部分,叫做这个一元一次不等式组的解集。 ⑶一元一次不等式组的解法:先解出各个不等式的解集,然后再找出它们的公共部分。 可以利用数轴来找。

例1:解不等式组,并将其解集在数轴上表示出来.23112.2x x x -

?-+-??, ①

≥ ②

例2:求不等式组中字母的取值:已知不等式组321

x x a +??-

1.(2012河北省)下列各数中,为不等式组???<->-040

32x x 解的是( )

A.-1 B.0 C.2 D.4 2.解不等式组

()6152432112

323x x x x ++???--?

?> ≥② ①

()461,315,x x x x +>-???-≤+?? ???

????

>+---+>--)

() (23

2)6

54

2(21225

69x x x x

(六)、列不等式(组)解应用题:

1. (2012陕西 )小宏准备用50元钱买甲、乙两种饮料共10瓶.已知甲饮料每瓶7元,乙饮料每瓶4元,则小宏最多能买几瓶甲饮料?

2. 某学校组织340名师生进行长途考察活动,带有行李170件,计划租用甲、乙两种型号的汽车共10辆.经了解,甲车每辆最多能载40人和16件行李,乙车每辆最多能载30人和20件行李.

⑴请你帮助学校设计所有可行的租车方案; ⑵如果甲车的租金为每辆2000元,乙车的租金为每辆1800元,问哪种可行方案使租车费用最省?

3.(2014年四川资阳)某商家计划从厂家采购空调和冰箱两种产品共20台,空调的采购单价y1(元/台)与采购数量x1(台)满足y1=﹣20x1+1500(0<x1≤20,x1为整数);冰箱的采购单价y2(元/台)与采购数量x2(台)满足y2=﹣10x2+1300(0<x2≤20,x2为整数).

(1)经商家与厂家协商,采购空调的数量不少于冰箱数量的,且空调采购单价不低于1200元,问该商家共有几种进货方案?

(2)该商家分别以1760元/台和1700元/台的销售单价售出空调和冰箱,且全部售完.在(1)的条件下,问采购空调多少台时总利润最大?并求最大利润.

能力提高

一、选择题

1.(2012山东泰安)将不等式组

841

163

x x

x x

+<-

?

?

≤-

?

的解集在数轴上表示出来,正确的是()

A B C D

2.若不等式

x b

x a

-<

?

?

+>

?

的解集为2

A.-2,3 B.2,-3 C.3,-2 D.-3,2

3. 若不等式组

1+

240

x a

x

>

?

?

-

?≤

有解,则a的取值范围是

A .a ≤3

B .a <3

C .a <2

D .a ≤2

4.( 2014?广西)在等腰△ABC 中,AB =AC ,其周长为20cm ,则AB 边的取值范围是( )

A . 1cm <A

B <4cm

B . 5cm <AB <10cm

C . 4cm <AB <8cm

D . 4cm <AB <10cm

5.如果不等式 ?

??>

无解,那么m 的取值范围是( ) A .m >8 B .m ≥8 C .m <8 D .m ≤8

二、填空题

1.若关于x 的一元一次不等式组0

122x a x x ->??

->-?

无解,则a 的取值范围是( )

2、若关于x 、y 的二元一次方程组?

??-=+-=+221

32y x k y x 的解满足y x +﹥1,则k 的取值范围是 .

3. 不等式组21432x x x x +>??≤+?的解集是 .

4.不等式组??

?

??<-≤+4

21121x x 的整数解是 .

5.若不等式组{

3

x x m >>的解集是x>3,则m 的取值范围是______.

三、解不等式组,并把解集在数轴上表示出来.

四、解决问题

1. (2014福州市)现有A ,B 两种商品,买2件A 商品和1件B 商品用了90元,买3件A 商品和2件B 商品

共用了160元

(1)求A ,B 两种商品每件多少元?

(2)如果小亮准备购买A ,B 两种商品共10件,总费用不超过350元,且不低于300元,问有几种购买方案,

哪种方案费用最低?

2. (2012贵州黔西南州)某工厂计划生产A、B两种产品共10件,其生产成本和利润如下表.

(1)若工厂计划获利14万元,问A、B两种产品应Array分别生产多少件?

(2)若工厂计划投入资金不多于44万元,且获利多

于14万元,问工厂有哪几种生产方案?

(3)在(2)的条件下,哪种生产方案获利最大?并求出最大利润.

3.(2014?孝感)我市荸荠喜获丰收,某生产基地收获荸荠40吨.经市场调查,可采用批发、零售、加工销售三种销

售方式,这三种销售方式每吨荸荠的利润如下表:

销售方式批发零售加工销售

利润(百元/吨)12 22 30

设按计划全部售出后的总利润为y百元,其中批发量为x吨,且加工销售量为15吨.

(1)求y与x之间的函数关系式;

(2)若零售量不超过批发量的4倍,求该生产基地按计划全部售完荸荠后获得的最大利润.

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