2026年作业本浙江教育出版社六年级数学下册浙教版第47页答案
1. 不计算,在$◯$里填上“>”、“<”或“=”。
$15×\frac{3}{5}◯15$ $1.5×1.1◯1.1$ $\frac{3}{4}÷\frac{2}{5}◯\frac{3}{4}×\frac{2}{5}$
$15÷\frac{3}{5}◯15$ $0.9÷0.01◯0.9$ $0.1×10◯10÷0.1$

答案

<,>,>,>,>,<

解析

1. $15×\frac{3}{5}$:$\frac{3}{5}$<1,一个数乘以小于1的数,结果小于原数,所以$15×\frac{3}{5}<15$;
2. $1.5×1.1$:$1.5>1$,一个数乘以大于1的数,结果大于另一个乘数1.1,所以$1.5×1.1>1.1$;
3. $\frac{3}{4}÷\frac{2}{5}=\frac{3}{4}×\frac{5}{2}$,$\frac{5}{2}>\frac{2}{5}$,所以$\frac{3}{4}÷\frac{2}{5}>\frac{3}{4}×\frac{2}{5}$;
4. $15÷\frac{3}{5}$:$\frac{3}{5}$<1,一个数除以小于1的数,结果大于原数,所以$15÷\frac{3}{5}>15$;
5. $0.9÷0.01$,$0.01<1$,一个数除以小于1的数,结果大于原数,所以$0.9÷0.01>0.9$;
6. $0.1×10 = 1$,$10÷0.1 = 100$,所以$0.1×10<10÷0.1$;
2. $18×\frac{5}{6}=$ $10÷\frac{5}{9}=$ $\frac{3}{4}×12=$ $3 - 3÷4=$
$\frac{5}{9}÷3=$ $\frac{2}{3}÷\frac{3}{2}=$ $12÷\frac{3}{4}=$ $\frac{2}{5}+\frac{3}{5}÷\frac{1}{2}=$
$\frac{7}{10}×\frac{5}{21}=$ $\frac{4}{5}÷4=$ $\frac{5}{7}÷\frac{6}{7}=$ $(\frac{5}{6}-\frac{2}{3})×12=$

答案

1. $18×\frac{5}{6}$
$\begin{aligned}&18×\frac{5}{6}\\=&3×5\\=&15\end{aligned}$
2. $10÷\frac{5}{9}$
$\begin{aligned}&10÷\frac{5}{9}\\=&10×\frac{9}{5}\\=&2×9\\=&18\end{aligned}$
3. $\frac{3}{4}×12$
$\begin{aligned}&\frac{3}{4}×12\\=&3×3\\=&9\end{aligned}$
4. $3 - 3÷4$
$\begin{aligned}&3 - 3÷4\\=&3 - \frac{3}{4}\\=&\frac{12}{4} - \frac{3}{4}\\=&\frac{9}{4}\end{aligned}$
5. $\frac{5}{9}÷3$
$\begin{aligned}&\frac{5}{9}÷3\\=&\frac{5}{9}×\frac{1}{3}\\=&\frac{5}{27}\end{aligned}$
6. $\frac{2}{3}÷\frac{3}{2}$
$\begin{aligned}&\frac{2}{3}÷\frac{3}{2}\\=&\frac{2}{3}×\frac{2}{3}\\=&\frac{4}{9}\end{aligned}$
7. $12÷\frac{3}{4}$
$\begin{aligned}&12÷\frac{3}{4}\\=&12×\frac{4}{3}\\=&4×4\\=&16\end{aligned}$
8. $\frac{2}{5}+\frac{3}{5}÷\frac{1}{2}$
$\begin{aligned}&\frac{2}{5}+\frac{3}{5}÷\frac{1}{2}\\=&\frac{2}{5}+\frac{3}{5}×2\\=&\frac{2}{5}+\frac{6}{5}\\=&\frac{8}{5}\end{aligned}$
9. $\frac{7}{10}×\frac{5}{21}$
$\begin{aligned}&\frac{7}{10}×\frac{5}{21}\\=&\frac{1}{2}×\frac{1}{3}\\=&\frac{1}{6}\end{aligned}$
10. $\frac{4}{5}÷4$
$\begin{aligned}&\frac{4}{5}÷4\\=&\frac{4}{5}×\frac{1}{4}\\=&\frac{1}{5}\end{aligned}$
11. $\frac{5}{7}÷\frac{6}{7}$
$\begin{aligned}&\frac{5}{7}÷\frac{6}{7}\\=&\frac{5}{7}×\frac{7}{6}\\=&\frac{5}{6}\end{aligned}$
12. $(\frac{5}{6}-\frac{2}{3})×12$
$\begin{aligned}&(\frac{5}{6}-\frac{2}{3})×12\\=&(\frac{5}{6}-\frac{4}{6})×12\\=&\frac{1}{6}×12\\=&2\end{aligned}$
答案依次为:15;18;9;$\frac{9}{4}$;$\frac{5}{27}$;$\frac{4}{9}$;16;$\frac{8}{5}$;$\frac{1}{6}$;$\frac{1}{5}$;$\frac{5}{6}$;2
3. 计算下面各题。
$4×\frac{4}{5}$ $(\frac{4}{7}-\frac{5}{21})×42$ $\frac{5}{16}÷\frac{3}{4}$
$3÷\frac{3}{4}×\frac{3}{4}÷3$ $99÷4+\frac{1}{4}$ $\frac{7}{10}×15÷\frac{7}{100}$
$\frac{4}{7}÷\frac{4}{7}÷\frac{4}{7}$ $\frac{3}{5}×\frac{10}{27}×\frac{9}{14}$ $\frac{4}{5}÷\frac{4}{9}×\frac{5}{9}$

答案

1. $4×\frac{4}{5}=\frac{16}{5}$
2. $(\frac{4}{7}-\frac{5}{21})×42=\frac{4}{7}×42 - \frac{5}{21}×42=24 - 10=14$
3. $\frac{5}{16}÷\frac{3}{4}=\frac{5}{16}×\frac{4}{3}=\frac{5}{12}$
4. $3÷\frac{3}{4}×\frac{3}{4}÷3=3×\frac{4}{3}×\frac{3}{4}×\frac{1}{3}=1$
5. $99÷4+\frac{1}{4}=\frac{99}{4}+\frac{1}{4}=\frac{100}{4}=25$
6. $\frac{7}{10}×15÷\frac{7}{100}=\frac{7}{10}×15×\frac{100}{7}=150$
7. $\frac{4}{7}÷\frac{4}{7}÷\frac{4}{7}=1÷\frac{4}{7}=\frac{7}{4}$
8. $\frac{3}{5}×\frac{10}{27}×\frac{9}{14}=\frac{2}{9}×\frac{9}{14}=\frac{1}{7}$
9. $\frac{4}{5}÷\frac{4}{9}×\frac{5}{9}=\frac{4}{5}×\frac{9}{4}×\frac{5}{9}=1$
4. 在$◯$里填上“>”、“<”或“=”。($a$,$b$均不为零)
(1) $a×\frac{3}{8}=b×\frac{3}{5}$ $a◯ b$ (2) $a÷\frac{3}{8}=b÷\frac{3}{5}$ $a◯ b$
(3) $a×\frac{3}{8}=b÷\frac{3}{8}$ $a◯ b$ (4) $\frac{3}{8}÷ a=\frac{3}{5}÷ b$ $a◯ b$

答案

(1) >;(2) <;(3) >;(4) <

解析

(1) 根据 $a×\frac{3}{8}=b×\frac{3}{5}$,可得 $\frac{a}{b}=\frac{\frac{3}{5}}{\frac{3}{8}}=\frac{8}{5}$,所以 $a> b$。
(2) 根据 $a÷\frac{3}{8}=b÷\frac{3}{5}$,即 $a×\frac{8}{3}=b×\frac{5}{3}$,则 $\frac{a}{b}=\frac{\frac{5}{3}}{\frac{8}{3}}=\frac{5}{8}$,所以 $a< b$。
(3) 根据 $a×\frac{3}{8}=b÷\frac{3}{8}$,即 $a×\frac{3}{8}=b×\frac{8}{3}$,那么 $\frac{a}{b}=\frac{\frac{8}{3}}{\frac{3}{8}}=\frac{64}{9}$,所以 $a> b$。
(4) 根据 $\frac{3}{8}÷ a=\frac{3}{5}÷ b$,即 $\frac{3}{8a}=\frac{3}{5b}$,可得 $\frac{a}{b}=\frac{5}{8}$,所以 $a< b$。