2025年天府前沿七年级数学上册北师大版


注:目前有些书本章节名称可能整理的还不是很完善,但都是按照顺序排列的,请同学们按照顺序仔细查找。练习册 2025年天府前沿七年级数学上册北师大版 答案主要是用来给同学们做完题方便对答案用的,请勿直接抄袭。



《2025年天府前沿七年级数学上册北师大版》

第60页
1. 指出下列变化中所运用的运算律:
(1) $ 5 × (-2) = -2 × 5 $ ( );
(2) $ -\frac{5}{6} + \frac{1}{3} = \frac{1}{3} - \frac{5}{6} $ ( );
(3) $ 6 × (-4) × (-5) = 6 × [(-4) × (-5)] $ ( );
(4) $ 24 × (\frac{5}{24} + 2 \frac{1}{6}) = 24 × \frac{5}{24} + 24 × 2 \frac{1}{6} $ ( ).
答案:
(1)乘法交换律
(2)加法交换律
(3)乘法结合律
(4)乘法分配律
2. 若 $ a < 0 $,$ b < 0 $,则 $ ab $______ $ 0 $;
若 $ a > 0 $,$ b > 0 $,则 $ ab $______ $ 0 $;
若 $ a < 0 $,$ b > 0 $,则 $ ab $______ $ 0 $.
答案: > > <
1. (2025·编写)计算:
(1) $ (-1 \frac{1}{2}) × (-3 \frac{1}{4}) × \frac{2}{3} = $______;
(2) $ (\frac{1}{2} - \frac{1}{3}) × 12 = $______;
(3) $ (-8) × 4 × (-1) × (-3) = $______.
答案:
(1)$\frac{13}{4}$
(2)2
(3)-96
2. (2025·编写)指出下列变化中所运用的运算律:
(1) $ 3 × (-2) = -2 × 3 $ ( );
(2) $ -\frac{1}{3} + \frac{1}{2} = \frac{1}{2} - \frac{1}{3} $ ( );
(3) $ 3 × (-2) × (-5) = 3 × [(-2) × (-5)] $ ( );
(4) $ 48 × (\frac{7}{24} - 2 \frac{1}{12}) = 48 × \frac{7}{24} - 48 × 2 \frac{1}{12} $ ( ).
答案:
(1)乘法交换律
(2)加法交换律
(3)乘法结合律
(4)乘法分配律
3. (2025·编写)计算:
(1) $ 1 + (\frac{2}{9} - \frac{1}{4} + \frac{1}{18}) × (-36) = $______;
(2) $ (-2) × (-7) × (+5) × (-\frac{1}{7}) = $______;
(3) $ -\frac{13}{17} × 19 - \frac{13}{17} × 15 = $______.
答案:
(1)0
(2)-10
(3)-26
4. (2025·编写)计算 $ (\frac{1}{3} + \frac{1}{4} - \frac{1}{2}) × 12 $ 时,应该运用 ( )

A.加法交换律
B.乘法分配律
C.乘法交换律
D.乘法结合律
答案: B
5. (2025·编写)下列计算,运用乘法结合律的是 ( )

A.$ \frac{5}{3} × \frac{6}{7} = \frac{6}{7} × \frac{5}{3} $
B.$ \frac{11}{12} × \frac{4}{5} × \frac{5}{4} = \frac{11}{12} × (\frac{4}{5} × \frac{5}{4}) $
C.$ \frac{5}{7} × 13 - \frac{5}{7} × 6 = \frac{5}{7} × (13 - 6) $
D.$ \frac{3}{8} - \frac{4}{5} + \frac{5}{8} = (\frac{3}{8} + \frac{5}{8}) - \frac{4}{5} $
答案: B
6. (2025·编写)下列各式计算正确的是 ( )

A.$ (-3) × (-2) = -6 $
B.$ (-4) × (-3) × (-5) = -60 $
C.$ (-\frac{1}{5}) × (-10) × 0 = 2 $
D.$ -6 × (-3) × (-6) × (-\frac{1}{3}) = -36 $
答案: B

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