2022年沪科版七年级数学下册复习知识点总结大全 .pdf

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1、relationship, establishe d equivalent relationship 14, and subject: appli cation problem (4)-scores and percentage application problem review content overview answers scores, and percentage application problem of key is: according to meaning, (1) determine standard volume (units 1) (2) find associat

2、e volume rate corresponds to relationship, Then in-line sol ution. Category fraction multiplication word problem score Division applications engineeri ng problem problem XV, a subject: review of the measurement of the amount of capacity, measurement and units of measurement of common units of measur

3、ement and their significa nce i n rate 1, currency, length, area, volume, unit size, volume, weight and rate. (Omitted) 2, commonly use d time units and their relationships. (Slightly) with a measurement units Zhijian of of poly 1, and of method 2, and poly method 3, and of method and poly method of

4、 relationship measurement distance of method 1, and tool measurement 2, and estimates 16, and subject: geometry preliminary knowledge (1)-line and angle review content line, and segment, and Ray, and vertical, and parallel, and angle angl e of classification (slightly) 17, and subject: geometry prel

5、iminary knowledge (2)-plane graphi cs review conte nt triangle, and edges shaped, a nd round, and fan axisymmetric graphics perimeter and area combination graphics of area subject : Preliminary knowledge (3)-review of solid content category 1-d shapes are divided into: cylinder and cone 2, column is

6、 divided into: cuboi d, square 3, cone cone of the features of cuboids and cubes relationship between characteristics of circular cone is slightly solid surfa ce area and volume 1, size 2, table .和努力学习好数学知识数学是一门研究数量、结构、变化以及空间模型等概念的学科；数学解题的关键就是知识和方法；知识是锁眼，方法是钥匙。缺少哪个都不能打开题目这把锁；那么我们的数学学习也要针对这两点进行。一、掌握课

7、本知识内容及内涵数学知识是数学解题的基石。只有掌握了课本知识的内容，理解知识的内涵，才能更好地运用它来解决问题。二、多看例题数学有的概念、定理较抽象，我们可以通过例题，将已有的概念具体化，使自己对知识的理解更加深刻，更加透彻！看例题时，还要注意以下几点：1、看一道例题，解决一类问题。不能只看皮毛，不看内涵。我们看例题，要注意总结并掌握其解题方法， 建立起更宽的解题思路。 不能看一道题就只会一道题，只记题目答案不记方法， 这样看例题也就失去了它本来的意义。每看一道题目，就应理清解题思路，掌握解题方法，再遇到同类型的题目，我们就不在难了。既然有“授人以鱼，不如授人以渔” ，那么我们是不是也可以说“

8、要鱼不如要渔”呢！2、我们不仅要看例题还要会总结，总结题型、解题思路和方法。运用了哪些数学思想。最好把总结的写出来。以后复习时再看，就事半功倍了。3、会模仿，也要创新。 在看例题的解题时，首先想自己遇到这个题怎么做，然后看例题怎么解答的， 之后我们还要思考还有没有其它方法和思路。我们最后看哪种方法更简便。三、多做练习“多”讲的是题型多，不是题目数量多。不怕难题，就怕生题。题海战术不一定好，但是接触的题型多了， 总结的解题方法多了。 以后遇到相同类型的题目也就不怕了。四、心细，多思，善问，勤总结数学是严谨的 ，做题目时要细心， 一个符号之差， 题目的解就可能完全不一样了，遇到问题要多思考， 培养

9、自己的数学思维， 思考实在不会的， 我们就要问，去弄懂。在数学学习过程中，我们要会总结，还要勤总结。多总结知识内容，总结解题方法，解题思想。 一方面能够起到复习巩固的作用，另一方面能提高自己的自学能力。数学的四大思维体系：数形结合、函数思想、分类讨论、方程思想。第六章实数精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 1 页，共 17 页relationship, establishe d equivalent relationship 14, and subject: appli cation problem (4)-scores and per

10、centage application problem review content overview answers scores, and percentage application problem of key is: according to meaning, (1) determine standard volume (units 1) (2) find associate volume rate corresponds to relationship, Then in -line sol ution. Category fraction multiplication word p

11、roblem score Division applications engineeri ng problem problem XV, a subject: review of the measurement of the amount of capacity, measurement and units of measurement of common units of measurement and their significa nce i n rate 1, currency, length, area, volume, unit size, volume, weight and ra

12、te. (Omitted) 2, commonly use d time units and their relationships. (Slightly) with a measurement units Zhijian of of poly 1, and of method 2, and poly method 3, and of method and poly method of relationship measurement distance of method 1, and tool measurement 2, and estimates 16, and subject: geo

13、metry preliminary knowledge (1)-line and angle review content line, and segment, and Ray, and vertical, and parallel, and angle angl e of classification (slightly) 17, and subject: geometry preliminary knowledge (2)-plane graphi cs review conte nt triangle, and edges shaped, a nd round, and fan axis

14、ymmetric graphics perimeter and area combination graphics of area subject : Preliminary knowledge (3)-review of solid content category 1-d shapes are divided into: cylinder and cone 2, column is divided into: cuboi d, square 3, cone cone of the features of cuboids and cubes relationship between char

15、acteristics of circular cone is slightly solid surfa ce area and volume 1, size 2, table .和一、知识总结（一）平方根与立方根1、平方根（1）定义：一般地，如果一个数的平方等于a，那么这个数叫做a 的平方根，也叫做二次方根。（2）表示：非负数a 的平方根记作a，读作“正负根号a” ， （a叫做被开方数）（3）性质：正数的平方根有两个，且互为相反数； 0的平方根为0；负数的没有平方根。（4）开平方：求平方根的运算叫做开平方。、平方根是开平方的结果；、开平方与平方互为逆运算。2、算术平方根（1）定义：正数 a

16、的正的平方根a叫做 a的算术平方根， 0 的算术平方根是0。（2）性质：（1）一个数 a的算术平方根具有非负性 ； 即：a0 恒成立 。（2）正数的算术平方根只有1 个，且为正数； 0 的算术平方根是0；负数的没有算术平方根。3、立方根：（1）定义：一般地，如果一个数的立方等于a，那么这个数叫做a 的立方根，也叫做三次方根。（2）表示： a的立方根记作3a，读作“三次根号a” （a叫做被开方数，3叫根指数）（3）性质：正数的立方根是1 个正数；负数的立方根是1 个负数； 0 的立方根是0。（二）实数1、无理数：无限不循环的小数。（一个无理数与若干有理数之间的运算结果还是无理数）2、实数： 有理

17、数和无理数统称为实数。3、实数分类：（1）按定义分（略）（2）按正负性分（略）4、实数与数轴上的点一一对应。5、实数的 相反数、绝对值、倒数： （与有理数的相反数、绝对值、倒数意义类似）6、实数的运算 ：实数与有理数一样，可以进行加、减、乘、除、乘方运算，正数及零可以进行开平方运算，任意一个实数可以进行开立方运算，而且 有理数的运算法则和运算律对于实数仍然适用。7、实数大小 ： （1）正数 0 负数；（2）两个负数相比，绝对值大的反而小；绝对值小的反而大。 （3）数轴上不同的点表示的数，右边点表示的数总比左边的点表示的数大。实数比较大小的方法：作差法、平方法、作商法、倒数法、估值法222222

18、二、解题实用 1、1.4142121.73232.2365 2、aa2a2aaa3333a 3、abbabababa0b精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 2 页，共 17 页relationship, establishe d equivalent relationship 14, and subject: appli cation problem (4)-scores and percentage application problem review content overview answers scores, and perc

19、entage application problem of key is: according to meaning, (1) determine standard volume (units 1) (2) find associate volume rate corresponds to relationship, Then in -line sol ution. Category fraction multiplication word problem score Division applications engineeri ng problem problem XV, a subjec

20、t: review of the measurement of the amount of capacity, measurement and units of measurement of common units of measurement and their significa nce i n rate 1, currency, length, area, volume, unit size, volume, weight and rate. (Omitted) 2, commonly use d time units and their relationships. (Slightl

21、y) with a measurement units Zhijian of of poly 1, and of method 2, and poly method 3, and of method and poly method of relationship measurement distance of method 1, and tool measurement 2, and estimates 16, and subject: geometry preliminary knowledge (1)-line and angle review content line, and segm

22、ent, and Ray, and vertical, and parallel, and angle angl e of classification (slightly) 17, and subject: geometry preliminary knowledge (2)-plane graphi cs review conte nt triangle, and edges shaped, a nd round, and fan axisymmetric graphics perimeter and area combination graphics of area subject :

23、Preliminary knowledge (3)-review of solid content category 1-d shapes are divided into: cylinder and cone 2, column is divided into: cuboi d, square 3, cone cone of the features of cuboids and cubes relationship between characteristics of circular cone is slightly solid surfa ce area and volume 1, s

24、ize 2, table .和- 3 - 三、典题练习1、16的平方根是；23-的算术平方根是；23-的立方根是。2、如果一个有理数的算术平方根与立方根相同，那么这个数是；如果一个有理数的平方根与立方根相同，那么这个数是。3、一个自然数的算术平方根是x，则与他相邻的下一个自然数的算术平方根是。4、下列各数中一定为正数的是（填序号）x 1x2x1x31x5、当x-1时，2x，-x,3x-和x1的大小关系。6、比较下列各组数的大小2-23-21与75412与112533与71-21-4与7、2-7的绝对值为，相反数为，倒数为。8、已知3x，y为 4 的平方根，0 xy，求x+y的值。9、已知02-

25、3xy，求x2+y的平方根。10、 如果一个非负数的平方根为2a-1和a-5，则这个数是。11、a为5的整数部分，b为5的小数部分，则a+2b的值为。12、 若aa 2012-a-2011，试求22011-a的值。 （提示：找出题中的隐含条件）第七章一元一次不等式与不等式组一、知识总结（一）不等式及其性质1、不等式：（ 1）定义用“”( 或“” ) ，“” (或“” ) 等不等号表示大小关系的式子，叫做不等式 . 用“”表示不等关系的式子也是不等式. （ 2）不等式的解：能使不等式成立的未知数的值，叫做不等式的解。（ 3）不等式的解集：一般地，一个含有未知数的不等式的所有解，组成这个不等式的解

26、集。求不等式的解集的过程叫做解不等式。不等式的解集与不等式的解的区别： 解集是能使不等式成立的未知数的取值范围,是所有解的集合 , 而不等式的解是使不等式成立的未知数的值。二者的关系是： 解集包括解 , 所有的解组成了解集。精选学习资料 - - - - - - - - - 名师归纳总结 - - - - - - -第 3 页，共 17 页relationship, establishe d equivalent relationship 14, and subject: appli cation problem (4)-scores and percentage application prob

27、lem review content overview answers scores, and percentage application problem of key is: according to meaning, (1) determine standard volume (units 1) (2) find associate volume rate corresponds to relationship, Then in -line sol ution. Category fraction multiplication word problem score Division ap

28、plications engineeri ng problem problem XV, a subject: review of the measurement of the amount of capacity, measurement and units of measurement of common units of measurement and their significa nce i n rate 1, currency, length, area, volume, unit size, volume, weight and rate. (Omitted) 2, commonl

29、y use d time units and their relationships. (Slightly) with a measurement units Zhijian of of poly 1, and of method 2, and poly method 3, and of method and poly method of relationship measurement distance of method 1, and tool measurement 2, and estimates 16, and subject: geometry preliminary knowle

30、dge (1)-line and angle review content line, and segment, and Ray, and vertical, and parallel, and angle angl e of classification (slightly) 17, and subject: geometry preliminary knowledge (2)-plane graphi cs review conte nt triangle, and edges shaped, a nd round, and fan axisymmetric graphics perime

31、ter and area combination graphics of area subject : Preliminary knowledge (3)-review of solid content category 1-d shapes are divided into: cylinder and cone 2, column is divided into: cuboi d, square 3, cone cone of the features of cuboids and cubes relationship between characteristics of circular

32、cone is slightly solid surfa ce area and volume 1, size 2, table .和（4）解不等式：求不等式解的过程叫做解不等式。2、不等式的基本性质性质 1：不等式的两边都加上( 或减去 ) 同一个整式，不等号的方向不变。即：如果ba，那么cbca. 性质 2：不等式的两边都乘上( 或除以 ) 同一个正数，不等号的方向不变。即：如果ba，并且0c，那么bcac；cbca. 性质 3：不等式的两边都乘上( 或除以 ) 同一个负数，不等号的方向改变 。即：如果ba，并且0c，那么bcac；cbca. 性质 4：如果ba，那么ab. （对称性）性质

33、 5：如果ba,cb, 那么ca. （传递性）（二）一元一次不等式1、定义：含有 一个未知数 ，未知数的 次数是 1，且不等号两边都是整式 的不等式，叫做一元一次不等式。2. 一元一次不等式的解法：根据是不等式的基本性质；一般步骤为：(1) 去分母； (2) 去括号； (3) 移项； (4)合并同类项；(5) 系数化为1. 解不等式应 注意： 去分母时，每一项都要乘同一个数，尤其不要漏乘常数项；移项时不要忘记变号；去括号时，若括号前面是负号，括号里的每一项都要变号；在不等式两边都乘( 或除以 ) 同一个负数时，不等号的方向要改变。3. 不等式的解集在数轴上表示：（ 1）边界：有等号的是实心圆圈

34、，无等号的是空心圆圈；（2） 方向：大向右，小向左（三）一元一次不等式组 1、定义： 有几个含有同一个未知数的一元一次不等式组成的不等式组，叫做一元一次不等式组 2、 （一元一次） 不等式组的解集：这几个不等式解集的公共部分，叫做这个 （一元一次）不等式组的解集。 3、解不等式组：求不等式组解集的过程，叫做解不等式组。 4、一元一次不等式组的解法 1 ）分别求出不等式组中各个不等式的解集 2 ）利用数轴求出这些不等式的解集的公共部分，即这个不等式组的解集。由两个一元一次不等式组成的不等式组的解集可归纳为下面四种情况：不等式组ba解集口诀记忆abxxbx同大取大axbxax同小取小精选学习资料

35、- - - - - - - - - 名师归纳总结 - - - - - - -第 4 页，共 17 页relationship, establishe d equivalent relationship 14, and subject: appli cation problem (4)-scores and percentage application problem review content overview answers scores, and percentage application problem of key is: according to meaning, (1) det

36、ermine standard volume (units 1) (2) find associate volume rate corresponds to relationship, Then in -line sol ution. Category fraction multiplication word problem score Division applications engineeri ng problem problem XV, a subject: review of the measurement of the amount of capacity, measurement

37、 and units of measurement of common units of measurement and their significa nce i n rate 1, currency, length, area, volume, unit size, volume, weight and rate. (Omitted) 2, commonly use d time units and their relationships. (Slightly) with a measurement units Zhijian of of poly 1, and of method 2,

38、and poly method 3, and of method and poly method of relationship measurement distance of method 1, and tool measurement 2, and estimates 16, and subject: geometry preliminary knowledge (1)-line and angle review content line, and segment, and Ray, and vertical, and parallel, and angle angl e of class

39、ification (slightly) 17, and subject: geometry preliminary knowledge (2)-plane graphi cs review conte nt triangle, and edges shaped, a nd round, and fan axisymmetric graphics perimeter and area combination graphics of area subject : Preliminary knowledge (3)-review of solid content category 1-d shap

40、es are divided into: cylinder and cone 2, column is divided into: cuboi d, square 3, cone cone of the features of cuboids and cubes relationship between characteristics of circular cone is slightly solid surfa ce area and volume 1, size 2, table .和- 5 - abxxbxa大小小大中间找abxx无解大大小小则无解（四）一元一次不等式（组）解决实际问题

41、解题的步骤：审题，找出不等关系设未知数列出不等式（组）求出不等式的解集找出符合题意的值作答。二、解题技巧一、有解无解问题：（1）abxxbbaa有解：无解：（2）axxbbbaa有解：无解：（3）abxxbbaa有解：无解：2、特征解问题：解题步骤：把原式中的要求的量（以下简记为m) 当作已知数，去解原式得到原式的解 （含m) 根据解的特征列出式子（关于m的式子 ) 解出m的值。例： 已知12axx的解集为1x，求a的值。解：解不等式12axx222222 把a当作已知数，去解原式得1xa222222 得到原式的解（含a)则11-a222222 根据解的特征列出式子解得2a222222 解出a

42、的值三、典题练习1、若关于x的不等式1x12mmx有解，则m的取值范围是？若无解呢？2、已知关于x，y的方程组myyx1x222的解满足0 xy，求m的取值范围。3、适当选择a 的取值范围，使1.7xa 的整数解：（1）x 只有一个整数解；（2）x 一个整数解也没有。4、解不等式（组）（1）322,352xxxx（ 2）. 3273, 4536,7342xxxxxx（3）.1)3(221,312233xxxxx（4）562x3 （ 5）.17)10(2383yyy5、若 m、n 为有理数，解关于x 的不等式 (m21)x n精选学习资料 - - - - - - - - - 名师归纳总结 - -

43、 - - - - -第 5 页，共 17 页relationship, establishe d equivalent relationship 14, and subject: appli cation problem (4)-scores and percentage application problem review content overview answers scores, and percentage application problem of key is: according to meaning, (1) determine standard volume (unit

44、s 1) (2) find associate volume rate corresponds to relationship, Then in -line sol ution. Category fraction multiplication word problem score Division applications engineeri ng problem problem XV, a subject: review of the measurement of the amount of capacity, measurement and units of measurement of

45、 common units of measurement and their significa nce i n rate 1, currency, length, area, volume, unit size, volume, weight and rate. (Omitted) 2, commonly use d time units and their relationships. (Slightly) with a measurement units Zhijian of of poly 1, and of method 2, and poly method 3, and of me

46、thod and poly method of relationship measurement distance of method 1, and tool measurement 2, and estimates 16, and subject: geometry preliminary knowledge (1)-line and angle review content line, and segment, and Ray, and vertical, and parallel, and angle angl e of classification (slightly) 17, and

47、 subject: geometry preliminary knowledge (2)-plane graphi cs review conte nt triangle, and edges shaped, a nd round, and fan axisymmetric graphics perimeter and area combination graphics of area subject : Preliminary knowledge (3)-review of solid content category 1-d shapes are divided into: cylinde

48、r and cone 2, column is divided into: cuboi d, square 3, cone cone of the features of cuboids and cubes relationship between characteristics of circular cone is slightly solid surfa ce area and volume 1, size 2, table .和6、已知关于x，y的方程组134,123pyxpyx的解满足xy，求p的取值范围。7、已知关于x的不等式组0 x542bx的整数解共有3 个，求b的取值范围。8

49、、已知 A2x23x2，B2x2 4x5，试比较A 与 B 的大小。9、已知 a 是自然数，关于x 的不等式组02,43xax的解集是x2，求 a 的值。10、某种商品进价为150 元，出售时标价为225 元，由于销售情况不好，商品准备降价出售，但要保证利润不低于10，那么商店最多降价多少元出售商品? 11、 某零件制造车间有20 名工人，已知每名工人每天可制造甲种零件6 个或乙种零件 5个，且每制造一个甲种零件可获利150 元，每制造一个乙种零件可获利260 元。在这 20名工人中，车间每天安排x名工人制造甲种零件，其余工人制造乙种零件。（1）若此车间每天所获利润为y(元) ，用x的代数式表

50、示y。（2）若要使每天所获利润不低于24000 元，至少要派多少名工人去制造乙种零件? 12、 某学校计划组织385 名师生租车旅游，现知道出租公司有42 座和 60 座客车， 42 座客车的租金为每辆320 元， 60 座客车的租金为每辆460 元。（1）若学校单独租用这两种客车各需多少钱? （2）若学校同时租用这两种客车8 辆(可以坐不满 )，而且比单独租用一种车辆节省租金，请选择最节省的租车方案。第八章整式乘除与因式分解一、知识总结（一）幂的运算：1、同底数幂乘法：同底数幂相乘，底数不变，指数相加。nmnmaaa2、同底数幂除法：同底数幂相除，底数不变，指数相减。nmnmaaa3、幂的乘