Question

11．計(jì)算 
7-1÷8+$\frac{1}{8}$×7       
（3$\frac{5}{7}$+1.63+2$\frac{2}{7}$+1.37 ）÷6$\frac{3}{4}$
（$\frac{3}{5}$+1$\frac{1}{4}$）×（1-$\frac{2}{5}$×2$\frac{1}{4}$）
[1$\frac{2}{9}$+（5.4-2$\frac{2}{3}$）×$\frac{5}{6}$]÷$\frac{1}{15}$．

Answer 1

分析 按照四則混合運(yùn)算的順序進(jìn)行計(jì)算即可，先計(jì)算括號(hào)內(nèi)部的再計(jì)算括號(hào)外面的，同級(jí)運(yùn)算從左向右依次計(jì)算．

解答 解：①7-1÷8+$\frac{1}{8}$×7
=7-$\frac{1}{8}$$+\frac{7}{8}$
=7$\frac{3}{4}$

       
②（3$\frac{5}{7}$+1.63+2$\frac{2}{7}$+1.37 ）÷6$\frac{3}{4}$
=（3$\frac{5}{7}$+1.63+2$\frac{2}{7}$+1.37 ）×$\frac{4}{27}$
=[（3$\frac{5}{7}$+2$\frac{2}{7}$）+（1.37+1.63）]×$\frac{4}{27}$
=[6+3]×$\frac{4}{27}$
=$\frac{4}{3}$


③（$\frac{3}{5}$+1$\frac{1}{4}$）×（1-$\frac{2}{5}$×2$\frac{1}{4}$）
=$\frac{37}{20}$×$\frac{1}{10}$
=$\frac{37}{200}$

④[1$\frac{2}{9}$+（5.4-2$\frac{2}{3}$）×$\frac{5}{6}$]÷$\frac{1}{15}$
=[$\frac{11}{9}$+（$\frac{27}{5}$$-\frac{8}{3}$）×$\frac{5}{6}$]×15
=[$\frac{11}{9}$+$\frac{41}{15}$×$\frac{5}{6}$]×15
=[$\frac{11}{9}$$+\frac{41}{18}$]×15
=$\frac{63}{18}$×15
=$\frac{105}{2}$

點(diǎn)評(píng) 考查了運(yùn)算定律與簡(jiǎn)便運(yùn)算，四則混合運(yùn)算．注意運(yùn)算順序和運(yùn)算法則，靈活運(yùn)用所學(xué)的運(yùn)算律簡(jiǎn)便計(jì)算．