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1. ★★★ 若 $1024 × 63 = p$，则 $1024 × 64$ 的值可表示为 （
A.$p+1$
B.$p+63$
C.$p+1024$
D.$\dfrac{63}{64}p$
C
）A.$p+1$
B.$p+63$
C.$p+1024$
D.$\dfrac{63}{64}p$
答案
因为1 024×63 = p,所以1 024×64 = 1 024×(63+1)= 1 024×63+1 024 = p+1 024. 故选C.
2. 对于两个数，$M=1\ 024× 10\ 251\ 025$，$N=1\ 025× 10\ 241\ 024$，则（
A.$M=N$
B.$M＞N$
C.$M＜N$
D.无法确定$M,N$的大小
A
）A.$M=N$
B.$M＞N$
C.$M＜N$
D.无法确定$M,N$的大小
答案
根据数的分成和乘法分配律,可得M=1 024×(10 250 000+1 025)= 1 024×10 250 000+1 024×1 025 =1 024×1 025×10 000+1 024×1 025 = 1 025×10 240 000+1 024×1 025 = 1 025×(10 240 000+1 024), N = 1 025×(10 240 000+1 024),所以M=N.故选A.
3. 计算: (1) $(-1000 - \dfrac{2}{9}) + (-1001 - \dfrac{4}{9}) + 2000 + \dfrac{5}{9} =$ ______;
(2) $124.68 + 324.68 + 524.68 + 724.68 + 924.68 =$ ______;
(3) $2023×(\dfrac{1}{1011} - \dfrac{1}{1012}) + 1011×(\dfrac{1}{1012} + \dfrac{1}{2023}) - 1012×(\dfrac{1}{1011} - \dfrac{1}{2023}) =$ ______.
(2) $124.68 + 324.68 + 524.68 + 724.68 + 924.68 =$ ______;
(3) $2023×(\dfrac{1}{1011} - \dfrac{1}{1012}) + 1011×(\dfrac{1}{1012} + \dfrac{1}{2023}) - 1012×(\dfrac{1}{1011} - \dfrac{1}{2023}) =$ ______.
答案
(1) $-\dfrac{10}{9}$ 【解析】$(-1\ 000-\dfrac{2}{9})+(-1\ 001-\dfrac{4}{9})+2\ 000+\dfrac{5}{9}=-1\ 000-\dfrac{2}{9}-1\ 001-\dfrac{4}{9}+2\ 000+\dfrac{5}{9}=(-1\ 000-1\ 001+2\ 000)+(-\dfrac{2}{9}-\dfrac{4}{9}+\dfrac{5}{9})=-1-\dfrac{1}{9}=-\dfrac{10}{9}.$
(2) 2 623.4 【解析】$124.68+324.68+524.68+724.68+924.68=(100+24.68)+(300+24.68)+(500+24.68)+(700+24.68)+(900+24.68)=100+300+500+700+900+24.68×5=2\ 500+24.68×5=2\ 623.4.$
(3) 1 【解析】$2\ 023×(\dfrac{1}{1\ 011}-\dfrac{1}{1\ 012})+1\ 011×(\dfrac{1}{1\ 012}+\dfrac{1}{2\ 023})-1\ 012×(\dfrac{1}{1\ 011}-\dfrac{1}{2\ 023})=(1\ 011+1\ 012)×(\dfrac{1}{1\ 011}-\dfrac{1}{1\ 012})+1\ 011×(\dfrac{1}{1\ 012}+\dfrac{1}{2\ 023})-1\ 012×(\dfrac{1}{1\ 011}-\dfrac{1}{2\ 023})=1\ 011×(\dfrac{1}{1\ 011}-\dfrac{1}{1\ 012})+1\ 012×(\dfrac{1}{1\ 011}-\dfrac{1}{1\ 012})+1\ 011×(\dfrac{1}{1\ 012}+\dfrac{1}{2\ 023})-1\ 012×(\dfrac{1}{1\ 011}-\dfrac{1}{2\ 023})=1-\dfrac{1\ 011}{1\ 012}+\dfrac{1\ 012}{1\ 011}-1+\dfrac{1\ 011}{1\ 012}+\dfrac{1\ 011}{2\ 023}-\dfrac{1\ 012}{1\ 011}+\dfrac{1\ 012}{2\ 023}=\dfrac{1\ 011}{2\ 023}+\dfrac{1\ 012}{2\ 023}=\dfrac{2\ 023}{2\ 023}=1.$
(2) 2 623.4 【解析】$124.68+324.68+524.68+724.68+924.68=(100+24.68)+(300+24.68)+(500+24.68)+(700+24.68)+(900+24.68)=100+300+500+700+900+24.68×5=2\ 500+24.68×5=2\ 623.4.$
(3) 1 【解析】$2\ 023×(\dfrac{1}{1\ 011}-\dfrac{1}{1\ 012})+1\ 011×(\dfrac{1}{1\ 012}+\dfrac{1}{2\ 023})-1\ 012×(\dfrac{1}{1\ 011}-\dfrac{1}{2\ 023})=(1\ 011+1\ 012)×(\dfrac{1}{1\ 011}-\dfrac{1}{1\ 012})+1\ 011×(\dfrac{1}{1\ 012}+\dfrac{1}{2\ 023})-1\ 012×(\dfrac{1}{1\ 011}-\dfrac{1}{2\ 023})=1\ 011×(\dfrac{1}{1\ 011}-\dfrac{1}{1\ 012})+1\ 012×(\dfrac{1}{1\ 011}-\dfrac{1}{1\ 012})+1\ 011×(\dfrac{1}{1\ 012}+\dfrac{1}{2\ 023})-1\ 012×(\dfrac{1}{1\ 011}-\dfrac{1}{2\ 023})=1-\dfrac{1\ 011}{1\ 012}+\dfrac{1\ 012}{1\ 011}-1+\dfrac{1\ 011}{1\ 012}+\dfrac{1\ 011}{2\ 023}-\dfrac{1\ 012}{1\ 011}+\dfrac{1\ 012}{2\ 023}=\dfrac{1\ 011}{2\ 023}+\dfrac{1\ 012}{2\ 023}=\dfrac{2\ 023}{2\ 023}=1.$
4. 计算：
(1) $9999×7778 + 3333×6666$；
(2) $765×64×0.5×2.5×0.125$。
(1) $9999×7778 + 3333×6666$；
(2) $765×64×0.5×2.5×0.125$。
答案
(1) $9\ 999×7\ 778+3\ 333×6\ 666=9\ 999×7\ 778+3\ 333×3×2\ 222=9\ 999×(7\ 778+2\ 222)=99\ 990\ 000.$
(2) $765×64×0.5×2.5×0.125=765×2×4×8×0.5×2.5×0.125=765×(2×0.5)×(4×2.5)×(8×0.125)=765×1×10×1=7\ 650.$
(2) $765×64×0.5×2.5×0.125=765×2×4×8×0.5×2.5×0.125=765×(2×0.5)×(4×2.5)×(8×0.125)=765×1×10×1=7\ 650.$
5. (2025·眉山期中)学习了有理数的乘法后,老师给同学们出了这样一道题:
计算:$49\dfrac{24}{25}×(-5)$,看谁算得又快又对.
有两位同学的解法如下.
小铭:原式$=-\dfrac{1249}{25}×5=-\dfrac{1249}{5}=-249\dfrac{4}{5}$.
小俊:原式$=(49+\dfrac{24}{25})×(-5)=49×(-5)+\_\_\_\_\_\_×(-5)=\_\_\_\_\_\_$.
(1)请补全小俊的解题过程.
(2)通过上面的解法对你的启发,你认为有其他的解法吗?如果有,请把它写出来.
(3)用最简便的方法计算:$19\dfrac{15}{16}×(-8)$.
＞＞进一步挑战进阶专题:P22 专题8~P25 专题11
计算:$49\dfrac{24}{25}×(-5)$,看谁算得又快又对.
有两位同学的解法如下.
小铭:原式$=-\dfrac{1249}{25}×5=-\dfrac{1249}{5}=-249\dfrac{4}{5}$.
小俊:原式$=(49+\dfrac{24}{25})×(-5)=49×(-5)+\_\_\_\_\_\_×(-5)=\_\_\_\_\_\_$.
(1)请补全小俊的解题过程.
(2)通过上面的解法对你的启发,你认为有其他的解法吗?如果有,请把它写出来.
(3)用最简便的方法计算:$19\dfrac{15}{16}×(-8)$.
＞＞进一步挑战进阶专题:P22 专题8~P25 专题11
答案
(1) $\dfrac{24}{25}$；$-249\dfrac{4}{5}$ 【解析】利用乘法分配律,原式$=(49+\dfrac{24}{25})×(-5)=49×(-5)+\dfrac{24}{25}×(-5)=-245-\dfrac{24}{5}=-245-4\dfrac{4}{5}=-249\dfrac{4}{5}.$
(2) 有其他的解法：$49\dfrac{24}{25}×(-5)=(50-\dfrac{1}{25})×(-5)=50×(-5)-\dfrac{1}{25}×(-5)=-250+\dfrac{1}{5}=-249\dfrac{4}{5}.$
(3) $19\dfrac{15}{16}×(-8)=(20-\dfrac{1}{16})×(-8)=20×(-8)-\dfrac{1}{16}×(-8)=-160+\dfrac{1}{2}=-159\dfrac{1}{2}.$
(2) 有其他的解法：$49\dfrac{24}{25}×(-5)=(50-\dfrac{1}{25})×(-5)=50×(-5)-\dfrac{1}{25}×(-5)=-250+\dfrac{1}{5}=-249\dfrac{4}{5}.$
(3) $19\dfrac{15}{16}×(-8)=(20-\dfrac{1}{16})×(-8)=20×(-8)-\dfrac{1}{16}×(-8)=-160+\dfrac{1}{2}=-159\dfrac{1}{2}.$
