一、填一填。
1. 在计算$\dfrac{5}{6}-\dfrac{3}{4}+\dfrac{1}{2}$时,应先算(
1. 在计算$\dfrac{5}{6}-\dfrac{3}{4}+\dfrac{1}{2}$时,应先算(
减
)法,再算(加
)法。答案
减,加
解析
分数加减混合运算按从左到右的顺序计算,先算减法,再算加法。
2. 在计算$\dfrac{1}{2}+(\dfrac{5}{6}-\dfrac{3}{4})$时,应先算(
减
)法,再算(加
)法。答案
减,加
解析
根据分数加减混合运算的运算顺序,有括号的先算括号里面的,再算括号外面的。所以在计算$\frac{1}{2}+(\frac{5}{6}-\frac{3}{4})$时,应先算括号里的减法,再算括号外的加法。
1. 一批大米,上午运走$\dfrac{1}{6}$,下午运走$\dfrac{1}{4}$,求还剩下几分之几没有运完,可列式(
A.$\dfrac{1}{6}+\dfrac{1}{4}$
B.$\dfrac{1}{6}-\dfrac{1}{4}$
C.$1-(\dfrac{1}{6}+\dfrac{1}{4})$
D.$\dfrac{1}{6}×\dfrac{1}{4}$
C
)。A.$\dfrac{1}{6}+\dfrac{1}{4}$
B.$\dfrac{1}{6}-\dfrac{1}{4}$
C.$1-(\dfrac{1}{6}+\dfrac{1}{4})$
D.$\dfrac{1}{6}×\dfrac{1}{4}$
答案
C
解析
把这批大米的总量看成单位“1”,用上午运走的分率加上下午运走的分率,可得到一共运走的分率,再用单位“1”减去一共运走的分率,就是剩下没运的分率,列式为$1 - (\frac{1}{6} + \frac{1}{4})$。
2. 如果,那么“?”所表示的图形是(



A.
B.
C.
D.
D
)。A.
B.
C.
D.
答案
D
解析
设整个圆为单位“1”。第一个图形是圆平均分成2份,阴影占1份,为$\frac{1}{2}$;第二个图形是圆平均分成3份,阴影占1份,为$\frac{1}{3}$。则“?”表示的分数为$1-\frac{1}{2}-\frac{1}{3}=\frac{6}{6}-\frac{3}{6}-\frac{2}{6}=\frac{1}{6}$,即圆平均分成6份,阴影占1份,对应选项D。
三、计算。(写出计算过程)
$\dfrac{9}{10}-(\dfrac{1}{5}+\dfrac{1}{2})$
$\dfrac{9}{10}-\dfrac{1}{5}+\dfrac{1}{2}$
$\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{5}{6}$
$\dfrac{1}{2}+(\dfrac{2}{3}+\dfrac{5}{6})$
$\dfrac{3}{10}+(\dfrac{2}{5}-\dfrac{1}{3})$
$\dfrac{7}{10}-(\dfrac{5}{8}-\dfrac{1}{4})$
$\dfrac{9}{10}-(\dfrac{1}{5}+\dfrac{1}{2})$
$\dfrac{9}{10}-\dfrac{1}{5}+\dfrac{1}{2}$
$\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{5}{6}$
$\dfrac{1}{2}+(\dfrac{2}{3}+\dfrac{5}{6})$
$\dfrac{3}{10}+(\dfrac{2}{5}-\dfrac{1}{3})$
$\dfrac{7}{10}-(\dfrac{5}{8}-\dfrac{1}{4})$
答案
三、计算。(写出计算过程)
1. $\dfrac{9}{10} - (\dfrac{1}{5} + \dfrac{1}{2})$
$=\dfrac{9}{10} - (\dfrac{2}{10} + \dfrac{5}{10})$
$=\dfrac{9}{10} - \dfrac{7}{10}$
$=\dfrac{2}{10} = \dfrac{1}{5}$
2. $\dfrac{9}{10} - \dfrac{1}{5} + \dfrac{1}{2}$
$=\dfrac{9}{10} - \dfrac{2}{10} + \dfrac{5}{10}$
$=\dfrac{7}{10} + \dfrac{5}{10}$
$=\dfrac{12}{10} = \dfrac{6}{5}$
3. $\dfrac{1}{2} + \dfrac{2}{3} + \dfrac{5}{6}$
$=\dfrac{3}{6} + \dfrac{4}{6} + \dfrac{5}{6}$
$=\dfrac{12}{6} = 2$
4. $\dfrac{1}{2} + (\dfrac{2}{3} + \dfrac{5}{6})$
$=\dfrac{1}{2} + (\dfrac{4}{6} + \dfrac{5}{6})$
$=\dfrac{1}{2} + \dfrac{9}{6}$
$=\dfrac{1}{2} + \dfrac{3}{2}$
$=\dfrac{4}{2} = 2$
5. $\dfrac{3}{10} + (\dfrac{2}{5} - \dfrac{1}{3})$
$=\dfrac{3}{10} + (\dfrac{6}{15} - \dfrac{5}{15})$
$=\dfrac{3}{10} + \dfrac{1}{15}$
$=\dfrac{9}{30} + \dfrac{2}{30}$
$=\dfrac{11}{30}$
6. $\dfrac{7}{10} - (\dfrac{5}{8} - \dfrac{1}{4})$
$=\dfrac{7}{10} - (\dfrac{5}{8} - \dfrac{2}{8})$
$=\dfrac{7}{10} - \dfrac{3}{8}$
$=\dfrac{28}{40} - \dfrac{15}{40}$
$=\dfrac{13}{40}$
1. $\dfrac{9}{10} - (\dfrac{1}{5} + \dfrac{1}{2})$
$=\dfrac{9}{10} - (\dfrac{2}{10} + \dfrac{5}{10})$
$=\dfrac{9}{10} - \dfrac{7}{10}$
$=\dfrac{2}{10} = \dfrac{1}{5}$
2. $\dfrac{9}{10} - \dfrac{1}{5} + \dfrac{1}{2}$
$=\dfrac{9}{10} - \dfrac{2}{10} + \dfrac{5}{10}$
$=\dfrac{7}{10} + \dfrac{5}{10}$
$=\dfrac{12}{10} = \dfrac{6}{5}$
3. $\dfrac{1}{2} + \dfrac{2}{3} + \dfrac{5}{6}$
$=\dfrac{3}{6} + \dfrac{4}{6} + \dfrac{5}{6}$
$=\dfrac{12}{6} = 2$
4. $\dfrac{1}{2} + (\dfrac{2}{3} + \dfrac{5}{6})$
$=\dfrac{1}{2} + (\dfrac{4}{6} + \dfrac{5}{6})$
$=\dfrac{1}{2} + \dfrac{9}{6}$
$=\dfrac{1}{2} + \dfrac{3}{2}$
$=\dfrac{4}{2} = 2$
5. $\dfrac{3}{10} + (\dfrac{2}{5} - \dfrac{1}{3})$
$=\dfrac{3}{10} + (\dfrac{6}{15} - \dfrac{5}{15})$
$=\dfrac{3}{10} + \dfrac{1}{15}$
$=\dfrac{9}{30} + \dfrac{2}{30}$
$=\dfrac{11}{30}$
6. $\dfrac{7}{10} - (\dfrac{5}{8} - \dfrac{1}{4})$
$=\dfrac{7}{10} - (\dfrac{5}{8} - \dfrac{2}{8})$
$=\dfrac{7}{10} - \dfrac{3}{8}$
$=\dfrac{28}{40} - \dfrac{15}{40}$
$=\dfrac{13}{40}$
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