(1)1.03÷0.3,商是3.4時(shí),余數(shù)是
 

(2)4700÷400,商是11時(shí),余數(shù)是
 

(3)被除數(shù)擴(kuò)大到它的10倍,除數(shù)縮小到它的10倍,商
 

(4)被除數(shù)縮小到它的10倍,除數(shù)
 
,商擴(kuò)大10倍.
(5)從10里減去
 
個(gè)0.4后,還剩下4.
(6)一個(gè)三位小數(shù)取近似值是0.80,這個(gè)小數(shù)在取近似值前最大可能是
 
,最小可能是
 

(7)兩個(gè)數(shù)相除的商是87.9,如果被除數(shù)和除數(shù)都擴(kuò)大20倍,那么所得的商是
 
分析:①根據(jù)有余數(shù)的除法的驗(yàn)算方法“被除數(shù)=商×除數(shù)+余數(shù)”可得:“被除數(shù)-商×除數(shù)=余數(shù)”,根據(jù)這一等量關(guān)系式即可算出答案;
②根據(jù):“被除數(shù)-商×除數(shù)=余數(shù)”,即可得出結(jié)論.
③商不變的規(guī)律:在除法算式里,被除數(shù)和除數(shù)同時(shí)擴(kuò)大或縮小相同的倍數(shù)(0除外),商不變;若被除數(shù)擴(kuò)大10倍,除數(shù)縮小10倍,那么商就會(huì)擴(kuò)大100倍;可舉例進(jìn)行驗(yàn)證;
④根據(jù)商不變的規(guī)律:在除法算式里,被除數(shù)和除數(shù)同時(shí)擴(kuò)大或縮小相同的倍數(shù)(0除外),商不變;若被除數(shù)縮小到它的10倍,要使商擴(kuò)大10倍,除數(shù)要縮小100倍;
⑤由于10-4=6,又6÷0.4=15,所以從10里面連續(xù)減去15個(gè)0.4后,還剩下4;
⑥四舍五入后是0.80的三位小數(shù),這個(gè)小數(shù)十分位最大是8,百分位上的數(shù)字是0,千分位上最大是4,該數(shù)最大是0.804;該數(shù)十分位上最小是7,百分位上最小是9,千分位上最小為5,該數(shù)最小是0.795;
⑦根據(jù)商不變的規(guī)律:在除法算式里,被除數(shù)和除數(shù)同時(shí)擴(kuò)大或縮小相同的倍數(shù)(0除外),商不變;由此可得:如果被除數(shù)和除數(shù)都擴(kuò)大20倍,那么所得的商不變,仍是87.9;據(jù)此解答.
解答:解:①1.03-3.4×0.3,
=1.03-1.02,
=0.01;

②4700-400×11,
=4700-4400,
=300;

③被除數(shù)擴(kuò)大到它的10倍,除數(shù)縮小到它的10倍,商擴(kuò)大100倍;
如:20÷10=2,被除數(shù)擴(kuò)大到它的10倍是200,除數(shù)縮小到它的10倍是1,
商是:200÷1=200,擴(kuò)大了100倍;

④被除數(shù)縮小到它的10倍,除數(shù)縮小100倍,商擴(kuò)大10倍;

⑤(10-4)÷0.4,
=6÷0.4,
=15;

⑥由分析可知:這個(gè)小數(shù)在取近似值前最大可能是0.804,最小可能是0.795;

⑦根據(jù)商不變的規(guī)律可知:兩個(gè)數(shù)相除的商是87.9,如果被除數(shù)和除數(shù)都擴(kuò)大20倍,那么所得的商仍是87.9;
故答案為:0.01,300,擴(kuò)大100倍,縮小100倍,15,0.804,0.795,87.9.
點(diǎn)評(píng):此題涉及的知識(shí)點(diǎn)較多,但都比較簡(jiǎn)單,屬于基礎(chǔ)題,只要認(rèn)真,容易完成,注意平時(shí)基礎(chǔ)知識(shí)的積累.
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