Question

11．計算、能簡便的要簡便．

①（3.75+4.1+2.35）×9.8	②150×9.9	③1234+2341+3412+4123
④4.6×3.7+54×0.37	⑤$\frac{9}{20}$÷[$\frac{1}{2}$×（$\frac{2}{5}$+$\frac{4}{5}$）]	⑥（$\frac{1}{51}$-$\frac{2}{87}$）÷$\frac{1}{29}$+22÷51

Answer 1

分析 ①括號里面留言加法的交換律解答；
②④⑥利用乘法的分配律解答；
③⑤根據(jù)整數(shù)，分數(shù)四則運算的順序進行解答．

解答 解：①（3.75+4.1+2.35）×9.8
=（3.75+2.35+4.1）×9.8
=（6.1+4.1）×9.8
=10.2×9.8
=（10+0.2）×9.8
=10×9.8+0.2×9.8
=98+1.96
=99.96

②150×9.9
=150×（10-0.1）
=150×10-0.1×150
=1500-15
=1485

③1234+2341+3412+4123
=3575+3412+4123
=6987+4123
=11110

④4.6×3.7+54×0.37
=4.6×3.7+5.4×3.7
=（4.6+5.4）×3.7
=10×3.7
=37．

⑤$\frac{9}{20}$÷[$\frac{1}{2}$×（$\frac{2}{5}$+$\frac{4}{5}$）]
=$\frac{9}{20}$÷[$\frac{1}{2}$×$\frac{6}{5}$]
=$\frac{9}{20}$÷$\frac{3}{5}$
=$\frac{3}{4}$

⑥（$\frac{1}{51}$-$\frac{2}{87}$）÷$\frac{1}{29}$+22÷51
=（$\frac{1}{51}$-$\frac{2}{87}$）×29+$\frac{22}{51}$
=$\frac{29}{51}$-$\frac{58}{87}$+$\frac{22}{51}$
=1-$\frac{58}{87}$
=$\frac{1}{3}$．

點評 本題主要考查了小數(shù)，分數(shù)的四則運算，注意根據(jù)題目的特點選擇合適的簡便方法．