《2025年预学与导学六年级数学上册人教版》

1. 计算。
（1）$\frac{1}{2} × \frac{3}{4}$
（2）$(\frac{2}{5} × \frac{1}{3}) × \frac{3}{4}$
（3）$(\frac{1}{2} + \frac{1}{4}) × \frac{1}{3}$
$\frac{3}{4} × \frac{1}{2}$
$\frac{2}{5} × (\frac{1}{3} × \frac{3}{4})$
$\frac{1}{2} × \frac{1}{3} + \frac{1}{4} × \frac{1}{3}$

答案： （1）$\frac{1}{2} × \frac{3}{4} = \frac{1×3}{2×4} = \frac{3}{8}$
$\frac{3}{4} × \frac{1}{2} = \frac{3×1}{4×2} = \frac{3}{8}$
（2）$(\frac{2}{5} × \frac{1}{3}) × \frac{3}{4} = \frac{2×1}{5×3} × \frac{3}{4} = \frac{2}{15} × \frac{3}{4} = \frac{2×3}{15×4} = \frac{6}{60} = \frac{1}{10}$
$\frac{2}{5} × (\frac{1}{3} × \frac{3}{4}) = \frac{2}{5} × \frac{1×3}{3×4} = \frac{2}{5} × \frac{1}{4} = \frac{2×1}{5×4} = \frac{2}{20} = \frac{1}{10}$
（3）$(\frac{1}{2} + \frac{1}{4}) × \frac{1}{3} = (\frac{2}{4} + \frac{1}{4}) × \frac{1}{3} = \frac{3}{4} × \frac{1}{3} = \frac{3×1}{4×3} = \frac{1}{4}$
$\frac{1}{2} × \frac{1}{3} + \frac{1}{4} × \frac{1}{3} = \frac{1}{6} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4}$

2. 把第1题中结果相等的算式用等号连接起来，并概括出运算定律。
例：$14 × 50 = 50 × 14$ 乘法交换律：$a × b = b × a$
$8×\frac{3}{4}×\frac{1}{2} = \frac{3}{4}×8×\frac{1}{2}$乘法交换律：$a× b = b× a$
$\frac{5}{7} × \frac{3}{5} × \frac{2}{3} = \frac{5}{7} ×(\frac{3}{5} × \frac{2}{3})$乘法结合律：$(a× b)× c = a×(b× c)$
$(\frac{1}{5} + \frac{2}{3}) × \frac{1}{2} = \frac{1}{5} × \frac{1}{2} + \frac{2}{3} × \frac{1}{2}$乘法分配律：$(a + b) × c = a × c + b × c$

答案： 解析：题目考查乘法交换律、乘法结合律以及乘法分配律的形式及表达。
答案：
$8×\frac{3}{4}×\frac{1}{2} = \frac{3}{4}×8×\frac{1}{2}$；乘法交换律：$a× b = b× a$；
$\frac{5}{7} × \frac{3}{5} × \frac{2}{3} = \frac{5}{7} ×(\frac{3}{5} × \frac{2}{3})$；乘法结合律：$(a× b)× c = a×(b× c)$；
$(\frac{1}{5} + \frac{2}{3}) × \frac{1}{2} = \frac{1}{5} × \frac{1}{2} + \frac{2}{3} × \frac{1}{2}$；乘法分配律：$(a + b) × c = a × c + b × c$。