5. (1) $465+47$
$=(100×\_\_\_\_\_\_+10×\_\_\_\_\_\_+1×\_\_\_\_\_\_)+(10×\_\_\_\_\_\_+1×\_\_\_\_\_\_)$
$=100×\_\_\_\_\_\_+10×(\_\_\_\_\_\_+\_\_\_\_\_\_)+1×(\_\_\_\_\_\_+\_\_\_\_\_\_)$
$=100×\_\_\_\_\_\_+10×\_\_\_\_\_\_+1×\_\_\_\_\_\_$
$=\_\_\_\_\_\_$
(2) $42×15$
$=40×(\_\_\_\_\_\_)+40×(\_\_\_\_\_\_)+2×(\_\_\_\_\_\_)+2×(\_\_\_\_\_\_)$
$=(\_\_\_\_\_\_)+(\_\_\_\_\_\_)+(\_\_\_\_\_\_)+(\_\_\_\_\_\_)$
$=(\_\_\_\_\_\_)$
$=(100×\_\_\_\_\_\_+10×\_\_\_\_\_\_+1×\_\_\_\_\_\_)+(10×\_\_\_\_\_\_+1×\_\_\_\_\_\_)$
$=100×\_\_\_\_\_\_+10×(\_\_\_\_\_\_+\_\_\_\_\_\_)+1×(\_\_\_\_\_\_+\_\_\_\_\_\_)$
$=100×\_\_\_\_\_\_+10×\_\_\_\_\_\_+1×\_\_\_\_\_\_$
$=\_\_\_\_\_\_$
(2) $42×15$
$=40×(\_\_\_\_\_\_)+40×(\_\_\_\_\_\_)+2×(\_\_\_\_\_\_)+2×(\_\_\_\_\_\_)$
$=(\_\_\_\_\_\_)+(\_\_\_\_\_\_)+(\_\_\_\_\_\_)+(\_\_\_\_\_\_)$
$=(\_\_\_\_\_\_)$
答案
5. (1) $465+47$
$=(100×\underline{4}+10×\underline{6}+1×\underline{5})+(10×\underline{4}+1×\underline{7})$
$=100×\underline{4}+10×(\underline{6}+\underline{4})+1×(\underline{5}+\underline{7})$
$=100×\underline{5}+10×\underline{1}+1×\underline{2}$
$=\underline{512}$
(2) $42×15$
$=40×(\underline{10})+40×(\underline{5})+2×(\underline{10})+2×(\underline{5})$
$=(\underline{400})+(\underline{200})+(\underline{20})+(\underline{10})$
$=(\underline{630})$
$=(100×\underline{4}+10×\underline{6}+1×\underline{5})+(10×\underline{4}+1×\underline{7})$
$=100×\underline{4}+10×(\underline{6}+\underline{4})+1×(\underline{5}+\underline{7})$
$=100×\underline{5}+10×\underline{1}+1×\underline{2}$
$=\underline{512}$
(2) $42×15$
$=40×(\underline{10})+40×(\underline{5})+2×(\underline{10})+2×(\underline{5})$
$=(\underline{400})+(\underline{200})+(\underline{20})+(\underline{10})$
$=(\underline{630})$
6. 旺山动物园原来门票单价是 30 元/张。从今年 5 月 1 日起,门票价格降低$\frac{1}{3}$,降价后门票单价是多少元/张?
答案
30×(1 - $\frac{1}{3}$)
= 30×$\frac{2}{3}$
= 20(元/张)
答:降价后门票单价是20元/张。
= 30×$\frac{2}{3}$
= 20(元/张)
答:降价后门票单价是20元/张。
7. 一个分数的分子加1后等于$\frac{1}{2}$,分母加1后等于$\frac{1}{3}$。这个分数是多少?
答案
$\frac{3}{8}$
8. 找规律并填空。
$1+3=4=2^2$
$1+3+5=9=3^2$
$1+3+5+7=16=4^2$
……
$1+3+5+\dots+19=(\quad)=(\quad)^2$
$1+3+5+\dots+(\quad)=(\quad)=20^2$
$1+3=4=2^2$
$1+3+5=9=3^2$
$1+3+5+7=16=4^2$
……
$1+3+5+\dots+19=(\quad)=(\quad)^2$
$1+3+5+\dots+(\quad)=(\quad)=20^2$
答案
$1+3+5+\dots+19=\boldsymbol{100}=\boldsymbol{10}^2$
$1+3+5+\dots+\boldsymbol{39}=\boldsymbol{400}=20^2$
$1+3+5+\dots+\boldsymbol{39}=\boldsymbol{400}=20^2$
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