作者：silverlegend

近些年来，小学教辅书有个奇怪的趋势，喜欢在印刷上做文章，搞的花花绿绿的，把低幼绘本那套画风延伸过来。语文英语之类的文科这么折腾也就罢了，数学这么搞，就很让人费解。

出版社的意图，大约是试图营造一种轻松愉快的学习氛围，某种程度上抵消学科本身的枯燥。但是很遗憾，单就数学而言，它本身就没啥乐趣可言，对于厌学的小朋友，这种表面功夫，实属枉然。

不少小学奥数教材也迎合了这种恶趣味，这里点名批评高思课本，喧宾夺主本末倒置，简直辣眼睛，虽然它家的高思导引属于奥数习题集标杆级的存在。

今天晒的这套《奥数精讲与测试》，实属一股清流，朴实无华，内容扎实，别有一番风味。总主编熊斌，是跟单墫单大爷边上边下的大佬，有他坐镇，妥妥没坑。这个系列贯穿小学到高中，畅销十几年，口碑甚好。

扔几个狗东链接……

一套六本，封面板式统一，颜色略有区别……

封皮直接秀出既往限量，底气十足……

封底介绍了两位主编……

扉页……

版权页……

再版前言……

文末有熊大爷的签名……

前言，介绍了本书的定位跟编排体例……

目录……

晒完整一章，感受一下……

开篇介绍知识点跟重难点，人狠话不多，要言不烦……

接着直接进入例题环节。这个修订版与时俱进，部分例题跟习题提供了配套讲解视频，页面侧边有二维码，扫之，即可播放视频……

每章例题不算太多，只是点到为止，对得起“精讲”二字。重头戏在习题部分，根据难度，分为ABC三篇，难度层层递进……

部分习题也配有视频讲解……

书末附有参考答案，很是详尽……

感受一下视频讲解的画风，同样的朴实无华……

这套书单册只有薄薄的两百来页，排版紧凑，内容扎实，毫不注水，是一套优秀的小学奥数教程，整体难度高于单墫大爷主编的小蓝本，可以根据个人喜好进行选择。

就这样吧，感谢围观！

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危致湛3524超难超难奥数题!!!!!!!!!!!!! -
徐宣牵17722177158 ______ 解:设南京到某地的距离为S千米,计划速度为V千米/小时,那么原定时间就是(S/V)小时 由按计划速度行驶1小时后,减速百分之40继续前进,这样他就比原定时间迟到2小时可得关系式 V*1+(1-40%)*V*【(S/V)+2-1】=S 整理得4V=S,所以原定时间是4小时 又按计划速度行驶1小时后,再按原速行驶90千米,那么他到某地的时间就只比原定时间迟1小时 可得关系式 V*1+90=V*(4-1) V=45 ∴S=45*4=180千米 答:南京到某地的距离为180千米

危致湛3524小学四年级奥数题 -
徐宣牵17722177158 ______ 一、按规律填数. 1)64,48,40,36,34,( ) 2)8,15,10,13,12,11,( ) 3)1、4、5、8、9、( )、13、( )、( ) 4)2、4、5、10、11、( )、( ) 5)5,9,13,17,21,( ),( ) 二、等差数列 1.在等差数列3,12,21,30,39,48,…中912是第几个数? 2.求1至100内所有不能被5...

危致湛3524奥数难题 -
徐宣牵17722177158 ______ 1.甲跑完全程,乙跑50米,丙跑40米,乙的速度与丙的速度比为5:4,乙再跑60米,丙可跑60/(5/4)=48米,乙领先丙60-48=12米2.设甲有X,乙有440-X 列方程X/2=3/5(440-X) X=240 乙有440-240=200元3.设牛有X,羊有(160-X),列方程 X+30...

危致湛3524数学奥数难题. -
徐宣牵17722177158 ______ 1.X-2(1/5X+30)=24,其中X为原汽油,X=140 2.1/8X=1/6Y,X为大长方形,Y为小,所以X/Y=4/3 3.设X天完成,32/100X=16,求出X为50,再剩下的是50-16=34天 4.X-1/11X=Y,X/Y=X除以10/11X=11/10,11/10-1=1/10,所以增加1/10 5.1.2X+1.3X...

危致湛3524奥数难题 -
徐宣牵17722177158 ______ 1. 1-1/3-1/4-1/5-1/6=1/202. 209=1+208=2+207=3+206,有规律知后一项为奇数,故取2+207,为第(207-1)/2+1=104项3. n^3+(7n)^2=n^3+49n^2=n^2 *(n+49),需要n+49为平方数,故n=15、32...

危致湛3524难度大的奥数题 -
徐宣牵17722177158 ______ 假设有X只小鸡,每只小鸡每天吃掉1个单位的饲料,根据饲料总量不变,有20(X-75)=15(X+100)20X-1500=15X+150020X-15X=1500+15005X=3000 X=600 农夫夫妇现在共有小鸡600只 ============= 算术方法 如果不卖掉75只小鸡,反而买进100只小鸡,则饲料能维持的天数由原来的20天减小为15天,即每天的消耗量的比为(20/15) 所以小鸡的数量比应该是(20/15),增多了[(20/15)-1](100+75)÷[(20/15)-1]=525(只) 这是卖掉75只后的小鸡数量,所以现在的小鸡数量是525+75=600(只)

危致湛3524几道奥数难题
徐宣牵17722177158 ______ 1 28.25 2 100 (分子、分母一样) 3 100/5=20,100/25=4 有10+20+4=34 个0 4 乙x,甲0.75x,丙x+1 x+0.75x+x+1=1101 x=400 分别为300、400、401

危致湛3524数学奥数难题
徐宣牵17722177158 ______ 有一列数,第一个数是100,第二个数是90,从第三个数开始,每个数都是它前面两个数的平均数.第三十个数的整数部分是_96____.

危致湛3524奥林匹克数学难题
徐宣牵17722177158 ______ 含盐量与含糖量的比就是杯中盐和糖的质量比 经过一番折腾,1号杯中盐和糖的质量分别为100*1/4*2/7,100(1-1/2)+100*1/2*2/7,比为1:9 2号杯中盐和糖的质量分别为100*1/4*(1-2/7)*(1-1/7),100*1/2*(1-2/7)*(1-1/7),,比为1:2 3号杯中盐和糖的质量分别为100*(1-3/4)+100*1/4*(1-2/7)*1/7,100*1/2*(1-2/7)*1/7,比为76:5

危致湛3524难题奥数,很急! -
徐宣牵17722177158 ______ ① 8743/4=2185.75 即1000-2185皆符合.共2185-1000+1=1186个 ② 数字为2时:99-35+1=65 数字为3时:99-00+1=100 数字为4时:99-00+1=100 数字为5时:99-00+1=100 数字为6时:99-00+1=100 数字为7时:99-00+1=100 数字为8时:99-00+1=100 数字为9时:97-00+1=98 总计:763个.