题目内容
14.脱式计算①58-(11$\frac{2}{5}$+12$\frac{3}{5}$+18$\frac{1}{9}$)
②444$\frac{4}{5}$×1.25
③1.5÷[1$\frac{2}{3}$×(3$\frac{1}{3}$+1$\frac{1}{6}$)]
④$\frac{15}{47}$×(2$\frac{4}{9}$-$\frac{20}{21}$)÷1$\frac{5}{7}$.
分析 ①先算小括号里的加法,再算括号外的减法;
②把444$\frac{4}{5}$化成444+$\frac{4}{5}$,再运用乘法的分配律进行简算;
③先算小括号里的加法,再算中括号里的乘法,最后算括号外的除法;
④把除以1$\frac{5}{7}$化成乘以$\frac{7}{12}$,再运用乘法的分配律进行简算.
解答 解:①58-(11$\frac{2}{5}$+12$\frac{3}{5}$+18$\frac{1}{9}$)
=58-(24+18$\frac{1}{9}$)
=58-42$\frac{1}{9}$
=15$\frac{8}{9}$;
②444$\frac{4}{5}$×1.25
=(444+$\frac{4}{5}$)×1.25
=444×1.25+$\frac{4}{5}$×1.25
=555+1
=556;
③1.5÷[1$\frac{2}{3}$×(3$\frac{1}{3}$+1$\frac{1}{6}$)]
=1.5÷[1$\frac{2}{3}$×$\frac{9}{2}$]
=1.5÷$\frac{15}{2}$
=$\frac{1}{5}$;
④$\frac{15}{47}$×(2$\frac{4}{9}$-$\frac{20}{21}$)÷1$\frac{5}{7}$
=$\frac{15}{47}$×(2$\frac{4}{9}$-$\frac{20}{21}$)×$\frac{7}{12}$
=$\frac{15}{47}$×2$\frac{4}{9}$×$\frac{7}{12}$-$\frac{15}{47}$×$\frac{20}{21}$×$\frac{7}{12}$
=$\frac{385}{47×3×6}$-$\frac{150}{47×3×6}$
=$\frac{235}{846}$.
点评 考查学生对四则运算法则以及运算定律及运算性质的掌握情况.
| (125+7)×8 | 3.27+6.4+2.73+4.6 | 5000÷8÷125 |
| 25×44 | 6.45-0.58-1.42 | 99×39+39 |
| 7×7+9= | 12+12÷12= | 0÷16×9= | 13×6+70= | 4×5×5= | 100÷4×6= |
| 450÷5÷9= | 280+70= | 60÷12= | 8×50= | 6+6×6= | 1000÷20= |
| A. | 48 | B. | 32 | C. | 6 |