《2025年暑假衔接起跑线小升初数学》

四、下面各题怎样简便就怎样算。
$9\frac {5}{6}-2.16-(1.84-\frac {1}{6})$
$31.4×0.65+4\frac {1}{2}×3.14-3\frac {7}{50}$
$48×(\frac {11}{12}-\frac {3}{8}+\frac {2}{3})$
$(\frac {3}{4}+\frac {3}{4}×99)÷\frac {3}{4}$
$17×(\frac {4}{17}+\frac {1}{25})×25$
$\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}+... +\frac {1}{9×10}$

答案： $9\frac {5}{6}-2.16-(1.84-\frac {1}{6})=9\frac {5}{6}-2.16-1.84+\frac {1}{6}=(9\frac {5}{6}+\frac {1}{6})-(2.16+1.84)=10-4=6$ $31.4×0.65+4\frac {1}{2}×3.14-3\frac {7}{50}=3.14×(6.5+4.5-1)=3.14×10=31.4$ $48×(\frac {11}{12}-\frac {3}{8}+\frac {2}{3})=48×\frac {11}{12}-48×\frac {3}{8}+48×\frac {2}{3}=44-18+32=58$ $(\frac {3}{4}+\frac {3}{4}×99)÷\frac {3}{4}=\frac {3}{4}×(1+99)×\frac {4}{3}=100$ $17×(\frac {4}{17}+\frac {1}{25})×25=17×25×\frac {4}{17}+17×25×\frac {1}{25}=100+17=117$ $\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}+... +\frac {1}{9×10}=1-\frac {1}{2}+\frac {1}{2}-\frac {1}{3}+\frac {1}{3}+... +\frac {1}{9}-\frac {1}{10}=1-\frac {1}{10}=\frac {9}{10}$