《2025年小升初考试新题型新考法真题精选详解数学》

1. $1000 + 999 - 998 - 997 + 996 + 995 - 994 - 993+\cdots+104 + 103 - 102 - 101$

答案： 原式=(1000 + 999 - 998 - 997)+(996 + 995 - 994 - 993)+…+(104 + 103 - 102 - 101)=4×(1000 - 101 + 1)÷4=900

2. $\left(1 - \dfrac{18}{51}\right)+\left(3 - \dfrac{18}{51}\times2\right)+\left(5 - \dfrac{18}{51}\times3\right)+\cdots+\left(97 - \dfrac{18}{51}\times49\right)+\left(99 - \dfrac{18}{51}\times50\right)$

答案： 原式=(1 + 3 + 5 +…+ 97 + 99)-($\frac{18}{51}+\frac{18}{51}×2+…+\frac{18}{51}×49+\frac{18}{51}×50$)=(1 + 99)×50÷2-$\frac{18}{51}$×(1 + 50)×50÷2=2500 - 450=2050

3. $\left(10 - \dfrac{2}{55}\times1\right)+\left(9 - \dfrac{2}{55}\times2\right)+\left(8 - \dfrac{2}{55}\times3\right)+\cdots+\left(1 - \dfrac{2}{55}\times10\right)$

答案： 原式=(10 + 9 + 8 +…+ 1)-$\frac{2}{55}$×(1 + 2 + 3 +…+ 10)=(10 + 1)×10÷2×(1-$\frac{2}{55}$)=55×$\frac{53}{55}$=53

4. $76\times\left(\dfrac{1}{23}-\dfrac{1}{53}\right)+23\times\left(\dfrac{1}{53}+\dfrac{1}{76}\right)-53\times\left(\dfrac{1}{23}-\dfrac{1}{76}\right)$

答案： 原式=76×$\frac{1}{23}$-$\frac{1}{53}$×76 + 23×$\frac{1}{53}$+ 23×$\frac{1}{76}$- 53×$\frac{1}{23}$+ 53×$\frac{1}{76}$=$\frac{76 - 53}{23}$-$\frac{76 - 23}{53}$+$\frac{23 + 53}{76}$=1 - 1 + 1=1

5. $999.3 - 998.2 + 997.3 - 996.2+\cdots+3.3 - 2.2 + 1.3 - 0.2$

答案： 原式=(999.3 - 998.2)+(997.3 - 996.2)+…+(3.3 - 2.2)+(1.3 - 0.2)=1.1×1000÷2=550

6. $99\dfrac{1}{2}-98\dfrac{1}{3}+97\dfrac{1}{2}-96\dfrac{1}{3}+95\dfrac{1}{2}-94\dfrac{1}{3}+\cdots+1\dfrac{1}{2}-\dfrac{1}{3}$

答案： 原式=($99\frac{1}{2}-98\frac{1}{3}$)+($97\frac{1}{2}-96\frac{1}{3}$)+…+($1\frac{1}{2}-\frac{1}{3}$)=$1\frac{1}{6}$×100×$\frac{1}{2}$=$58\frac{1}{3}$

7. $1949\times\left(\dfrac{1}{43}-\dfrac{1}{1992}\right)+43\times\left(\dfrac{1}{1949}-\dfrac{1}{1992}\right)-1992\times\left(\dfrac{1}{1949}+\dfrac{1}{43}\right)+3$

答案： 原式=($\frac{1949}{43}-\frac{1949}{1992}$)+($\frac{43}{1949}-\frac{43}{1992}$)-($\frac{1992}{1949}+\frac{1992}{43}$)+3=3+($\frac{1949}{43}-\frac{1992}{43}$)-($\frac{1949}{1992}+\frac{43}{1992}$)+($\frac{43}{1949}-\frac{1992}{1949}$)=3 - 1 - 1 - 1=0

8. $\dfrac{262}{426}\times\left(\dfrac{628}{862}-\dfrac{420}{648}\right)+\dfrac{628}{862}\times\left(\dfrac{420}{648}-\dfrac{262}{426}\right)+\dfrac{420}{648}\times\left(\dfrac{262}{462}-\dfrac{628}{862}\right)$

答案： 原式=$\frac{262}{426}$×$\frac{628}{862}$-$\frac{262}{426}$×$\frac{420}{648}$+$\frac{628}{862}$×$\frac{420}{648}$-$\frac{628}{862}$×$\frac{262}{426}$+$\frac{420}{648}$×$\frac{262}{426}$-$\frac{420}{648}$×$\frac{628}{862}$=($\frac{262}{426}$×$\frac{628}{862}$-$\frac{628}{862}$×$\frac{262}{426}$)+($\frac{628}{862}$×$\frac{420}{648}$-$\frac{420}{648}$×$\frac{628}{862}$)+($\frac{420}{648}$×$\frac{262}{426}$-$\frac{262}{426}$×$\frac{420}{648}$)=0

9. $\left(1 - \dfrac{1}{18}\times1\right)+\left(1 - \dfrac{1}{9}\times1\right)+\left(3 - \dfrac{1}{18}\times3\right)+\left(4 - \dfrac{1}{9}\times2\right)+\left(5 - \dfrac{1}{18}\times5\right)+\left(6 - \dfrac{1}{9}\times3\right)$

答案： 原式=(1 + 1 + 3 + 4 + 5 + 6)-($\frac{1}{18}$×1+$\frac{1}{18}$×3+$\frac{1}{18}$×5+$\frac{1}{9}$×1+$\frac{1}{9}$×2+$\frac{1}{9}$×3)=20 - ($\frac{1}{2}+\frac{2}{3}$)=$18\frac{5}{6}$