《2025年拔尖特训七年级数学上册人教版》

3. 计算：
(1)$3\frac{1}{7}×\frac{21}{22}×(3\frac{1}{7}-7\frac{1}{3})×(-\frac{7}{22})$.
(2)$(-99\frac{22}{23})×(-69)+1.25×(-\frac{81}{20})×(-8)$.
(3)$(-\frac{1}{84})÷(\frac{2}{3}+\frac{3}{7}-\frac{5}{28}-\frac{1}{4})$.

答案： （1）原式$=\frac{22}{7}×\frac{21}{22}×(3\frac{1}{7}-$$7\frac{1}{3})×(-\frac{7}{22})=-\frac{22}{7}×\frac{21}{22}×$$(\frac{22}{7}-\frac{22}{3})×\frac{7}{22}=-3×(\frac{22}{7}×\frac{7}{22}-$$\frac{22}{3}×\frac{7}{22})=-3×(1-\frac{7}{3})=-3×$$(-\frac{4}{3})=4$.（2）原式$=(-100+\frac{1}{23})×(-69)+$$\frac{5}{4}×(-\frac{81}{20})×(-8)=6900-3+$$\frac{5}{4}×\frac{81}{20}×8=6900-3+\frac{81}{2}=$$6937.5$.（3）因为原式的倒数为$(\frac{2}{3}+\frac{3}{7}-$$\frac{5}{28}-\frac{1}{4})÷(-\frac{1}{84})=(\frac{2}{3}+\frac{3}{7}-$$\frac{5}{28}-\frac{1}{4})×(-84)=\frac{2}{3}×(-84)+$$\frac{3}{7}×(-84)-\frac{5}{28}×(-84)-\frac{1}{4}×$$(-84)=-56-36+15+21=-56$，所以原式$=-\frac{1}{56}$.

4. 计算：
(1)$25×\frac{3}{4}-(-25)×\frac{1}{2}+25×(-\frac{1}{4})$.
(2)$(-3.61)×0.75+0.61×\frac{3}{4}+(-0.2)×75\%$.
(3)$-14×\frac{2}{3}-0.35×\frac{2}{7}+\frac{1}{3}×(-14)-\frac{5}{7}×0.35$.
(4)$2\frac{1}{3}×\frac{7}{13}-(-\frac{7}{13})×(-14)+(-\frac{4}{3})÷\frac{13}{7}$.
(5)$0.7×19\frac{5}{9}+2\frac{3}{4}×(-14)+\frac{7}{10}×\frac{4}{9}-3.25×14$.
(6)$(-3.85)×(-13)+(-13)×(-6.15)+0.79×\frac{7}{15}+\frac{8}{15}×0.79$.

答案： （1）25.（2）-2.4.（3）原式$=-14×(\frac{2}{3}+\frac{1}{3})-0.35×$$(\frac{2}{7}+\frac{5}{7})=-14-0.35=-14.35$.（4）原式$=2\frac{1}{3}×\frac{7}{13}-\frac{7}{13}×14-$$\frac{4}{3}×\frac{7}{13}=\frac{7}{13}×(2\frac{1}{3}-14-\frac{4}{3})=$$\frac{7}{13}×(-13)=-7$.（5）原式$=0.7×(19\frac{5}{9}+\frac{4}{9})-$$14×(2\frac{3}{4}+3.25)=0.7×20-14×$$6=14-84=-70$.（6）原式$=13×(3.85+6.15)+$$0.79×(\frac{7}{15}+\frac{8}{15})=13×10+0.79×$$1=130.79$.