1. $21×\frac{16}{7}×\frac{7}{8}=□◯(□◯□)$,运用了(
乘法结合律
)。答案
1.$21× \left(\dfrac{16}{7}× \dfrac{7}{8}\right)$ 乘法结合律
2. $\frac{5}{8}×\frac{2}{3}×\frac{8}{15}=(□◯□)◯□$,运用了(
乘法交换律与乘法结合律
)。答案
2.$\left(\dfrac{5}{8}× \dfrac{8}{15}\right)× \dfrac{2}{3}$ 乘法交换律与乘法结合律
3. $(\frac{4}{7}+\frac{1}{4})×28=□◯□◯□◯□$,运用了(
乘法分配律
)。答案
3.$\dfrac{4}{7}× 28+\dfrac{1}{4}× 28$ 乘法分配律
4. $54×(\frac{8}{9}-\frac{5}{6})=□◯□◯□◯□$,运用了(
乘法分配律
)。答案
4.$54× \dfrac{8}{9}-54× \dfrac{5}{6}$ 乘法分配律
二、我会算。(怎样简便怎样算)
$\frac{7}{5}×\frac{9}{23}×\frac{5}{7}$ $\frac{7}{11}×(\frac{3}{4}×\frac{11}{5})$ $(8+\frac{4}{5})×\frac{5}{8}$ $\frac{7}{12}×\frac{3}{4}+\frac{1}{4}×\frac{7}{12}$
$\frac{7}{5}×\frac{9}{23}×\frac{5}{7}$ $\frac{7}{11}×(\frac{3}{4}×\frac{11}{5})$ $(8+\frac{4}{5})×\frac{5}{8}$ $\frac{7}{12}×\frac{3}{4}+\frac{1}{4}×\frac{7}{12}$
答案
二、$\dfrac{9}{23}$ $\dfrac{21}{20}$ $5\dfrac{1}{2}$ $\dfrac{7}{12}$
解析
$\frac{7}{5}×\frac{9}{23}×\frac{5}{7}=\frac{7}{5}×\frac{5}{7}×\frac{9}{23}=1×\frac{9}{23}=\frac{9}{23}$
$\frac{7}{11}×(\frac{3}{4}×\frac{11}{5})=\frac{7}{11}×\frac{11}{5}×\frac{3}{4}=\frac{7}{5}×\frac{3}{4}=\frac{21}{20}$
$(8+\frac{4}{5})×\frac{5}{8}=8×\frac{5}{8}+\frac{4}{5}×\frac{5}{8}=5+\frac{1}{2}=5\frac{1}{2}$
$\frac{7}{12}×\frac{3}{4}+\frac{1}{4}×\frac{7}{12}=\frac{7}{12}×(\frac{3}{4}+\frac{1}{4})=\frac{7}{12}×1=\frac{7}{12}$
$\frac{7}{11}×(\frac{3}{4}×\frac{11}{5})=\frac{7}{11}×\frac{11}{5}×\frac{3}{4}=\frac{7}{5}×\frac{3}{4}=\frac{21}{20}$
$(8+\frac{4}{5})×\frac{5}{8}=8×\frac{5}{8}+\frac{4}{5}×\frac{5}{8}=5+\frac{1}{2}=5\frac{1}{2}$
$\frac{7}{12}×\frac{3}{4}+\frac{1}{4}×\frac{7}{12}=\frac{7}{12}×(\frac{3}{4}+\frac{1}{4})=\frac{7}{12}×1=\frac{7}{12}$
三、我会解答。
一辆汽车平均每千米耗油$\frac{3}{25}$L。这辆汽车第一天行驶了47km,第二天行驶了53km,这两天一共耗油多少升?
一辆汽车平均每千米耗油$\frac{3}{25}$L。这辆汽车第一天行驶了47km,第二天行驶了53km,这两天一共耗油多少升?
答案
三、$\dfrac{3}{25}× (47+53)=\dfrac{3}{25}× 100=12(L)$
解析
$\dfrac{3}{25} × (47 + 53) = \dfrac{3}{25} × 100 = 12\ (L)$
★四、我会巧思妙算。
$\frac{7}{9}×\frac{14}{15}+\frac{7}{15}×\frac{1}{9}$ $26×28×(\frac{1}{26×27}+\frac{1}{27×28})$
$\frac{7}{9}×\frac{14}{15}+\frac{7}{15}×\frac{1}{9}$ $26×28×(\frac{1}{26×27}+\frac{1}{27×28})$
答案
四、$\dfrac{7}{9}× \dfrac{14}{15}+\dfrac{7}{15}× \dfrac{1}{9}$
$=\dfrac{7}{9}× \dfrac{14}{15}+\dfrac{7}{9}× \dfrac{1}{15}$
$=\dfrac{7}{9}× \left(\dfrac{14}{15}+\dfrac{1}{15}\right)$
$=\dfrac{7}{9}× 1$
$=\dfrac{7}{9}$
$26× 28× \left(\dfrac{1}{26× 27}+\dfrac{1}{27× 28}\right)$
$=26× 28× \dfrac{1}{26× 27}+26× 28× \dfrac{1}{27× 28}$
$=\dfrac{28}{27}+\dfrac{26}{27}$
$=\dfrac{54}{27}$
$=2$
$=\dfrac{7}{9}× \dfrac{14}{15}+\dfrac{7}{9}× \dfrac{1}{15}$
$=\dfrac{7}{9}× \left(\dfrac{14}{15}+\dfrac{1}{15}\right)$
$=\dfrac{7}{9}× 1$
$=\dfrac{7}{9}$
$26× 28× \left(\dfrac{1}{26× 27}+\dfrac{1}{27× 28}\right)$
$=26× 28× \dfrac{1}{26× 27}+26× 28× \dfrac{1}{27× 28}$
$=\dfrac{28}{27}+\dfrac{26}{27}$
$=\dfrac{54}{27}$
$=2$
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