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電子課本網(wǎng) 第105頁

第105頁

信息發(fā)布者:
$1 - \frac{3} {8} - \frac{3} {10} = \frac{13} {40}$
$答:還剩下宣紙總量的\frac {13}{40}.$
$10分=\frac{1}{6}時$
$\ \frac{1} {3}+\frac{1} {6}=\frac{1} {2}(時)$
$\ 2÷8=\frac{1}{4}\ $
$1 - \frac{1} {2} - \frac{1} {4} = \frac{1} {4}$
$答:剩下的路程占全程的\frac {1}{4}$
2
$\frac{2}{9}$
$1-{\frac{2018} {2021}} = {\frac{3} {2021}}\ $
$1-{\frac{2020} {2023}} = {\frac{3} {2023}}\ $
$1-{\frac{2022} {2025}} = \frac{3} {2025}$
$\ 因為\frac{3} {2021} > \frac{3} {2023} > \frac{3} {2025},$
$所以\frac{2018} {2021}<\frac{2020} {2023} < \frac{2022} {2025}。$
$100÷2+1=51(棵) ,51×2=102(棵)\ $
$51÷ (2+1)=17(組)\ $
$香樟樹:17×2×2=68(棵),68÷102=\frac{2}{3}\ $
$梧桐樹:17×1×2=34(棵),34÷102=\frac{1}{3}$
$答:香樟樹占\frac {2}{3},梧桐樹占\frac {1}{3}。$
$\frac{48} {49}$
$\frac{4}{9}$
$2△4=\frac {1}{2}+\frac {1}{4}=\frac {3}{4}$
$(2△4)○(\frac {2}{3}+\frac {1}{5})$
$=\frac {3}{4}○(\frac {2}{3}+\frac {1}{5})$
$=(\frac {2}{3}+\frac {1}{5})-\frac {3}{4}$
$=\frac {7}{60}$