《2025年苏州精品六升七数学》

1. 将$4x + 8错写成4(x + 8)$后,结果比原来( )。

A.多 4
B.少 4
C.多 24
D.少 6

答案： C

2. 在估算$5.01×7.99$时,看成( )的误差较小。

A.$6×8$
B.$5×7$
C.$5×8$
D.$6×7$

答案： C

3. 在计算$15×12$时,聪聪使用的方法是$15×4×3$,下面的点子图中,( )能表示这种思路。

答案： C

4. 为了确保通信安全,信息需要加密再传输。现规定加密的规则：明文$(a,b)加密变成密文后是(4a + 3b,4b^{2} - 2)$。如：明文$(2,4)$,密文是$(20,62)$。密文$(17,34)$的明文是( )。

A.$(3,2)$
B.$(1,3)$
C.$(2,3)$
D.$(3,1)$

答案： C

5. 小马虎把$7.25 - (m + 2.25)算成了7.25 - m + 2.25$。他计算的结果比正确的结果( )。

A.多 2.25
B.少 2.25
C.多 4.5
D.少 4.5

答案： C

三、用简便方法计算下面各题。
$\frac{2023}{2024}×2025$ $56÷(56 + \frac{56}{57})$
$1111×6666 + 7778×3333$
$(1 + \frac{1}{2})×(1 + \frac{1}{3})×(1 + \frac{1}{4})×…×(1 + \frac{1}{100})$
$\frac{5}{6} - \frac{7}{12} + \frac{9}{20} - \frac{11}{30} + \frac{13}{42} - \frac{15}{56}$
$\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \frac{1}{30} + \frac{1}{42}$
$\frac{1}{2} + \frac{5}{6} + \frac{11}{12} + \frac{19}{20} + \frac{29}{30} + \frac{41}{42}$
$\frac{3}{2} + \frac{7}{6} + \frac{13}{12} + \frac{21}{20} + \frac{31}{30} + \frac{43}{42}$
$(\frac{4}{7}×\frac{1}{9}×\frac{4}{11})÷(\frac{2}{11}×\frac{2}{7}×\frac{5}{9})$
$\frac{2022×2023 - 1}{2022 + 2021×2023}$
$\frac{1}{21} + \frac{202}{2121} + \frac{50505}{212121} + \frac{13131313}{21212121}$

答案： $\frac{2023}{2024}×2025 = \frac{2023}{2024}×(2024 + 1) = \frac{2023}{2024}×2024 + \frac{2023}{2024}×1 = 2023\frac{2023}{2024}$
56÷（56 + $\frac{56}{57}$） = （56÷56）÷[（56 + $\frac{56}{57}$）÷56] = 1÷（1 + $\frac{1}{57}$） = 1÷$\frac{58}{57}$ = $\frac{57}{58}$　1111×6666 + 7778×3333 = 3333×2222 + 7778×3333 = （2222 + 7778）×3333 = 10000×3333 = 33330000　$(1 + \frac{1}{2})×(1 + \frac{1}{3})×(1 + \frac{1}{4})×\cdots×(1 + \frac{1}{100}) = \frac{3}{2}×\frac{4}{3}×\frac{5}{4}×\cdots×\frac{101}{100} = \frac{101}{2}$　$\frac{5}{6} - \frac{7}{12} + \frac{9}{20} - \frac{11}{30} + \frac{13}{42} - \frac{15}{56} = (\frac{1}{2} + \frac{1}{3}) - (\frac{1}{3} + \frac{1}{4}) + (\frac{1}{4} + \frac{1}{5}) - (\frac{1}{5} + \frac{1}{6}) + (\frac{1}{6} + \frac{1}{7}) - (\frac{1}{7} + \frac{1}{8}) = \frac{1}{2} + \frac{1}{3} - \frac{1}{3} - \frac{1}{4} + \frac{1}{4} + \frac{1}{5} - \frac{1}{5} - \frac{1}{6} + \frac{1}{6} + \frac{1}{7} - \frac{1}{7} - \frac{1}{8} = \frac{1}{2} - \frac{1}{8} = \frac{3}{8}$　$\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \frac{1}{30} + \frac{1}{42} = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + \frac{1}{5} - \frac{1}{6} + \frac{1}{6} - \frac{1}{7} = 1 - \frac{1}{7} = \frac{6}{7}$
$\frac{1}{2} + \frac{5}{6} + \frac{11}{12} + \frac{19}{20} + \frac{29}{30} + \frac{41}{42} = 1×6 - (\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \frac{1}{30} + \frac{1}{42}) = 6 - (1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + \frac{1}{5} - \frac{1}{6} + \frac{1}{6} - \frac{1}{7}) = 6 - (1 - \frac{1}{7}) = 5\frac{1}{7}$
$\frac{3}{2} + \frac{7}{6} + \frac{13}{12} + \frac{21}{20} + \frac{31}{30} + \frac{43}{42} = 1×6 + (\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \frac{1}{30} + \frac{1}{42}) = 6 + (1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + \frac{1}{5} - \frac{1}{6} + \frac{1}{6} - \frac{1}{7}) = 6 + (1 - \frac{1}{7}) = 6\frac{6}{7}$
$(\frac{4}{7}×\frac{1}{9}×\frac{4}{11})÷(\frac{2}{11}×\frac{2}{7}×\frac{5}{9}) = (\frac{4}{7}÷\frac{2}{7})×(\frac{1}{9}÷\frac{5}{9})×(\frac{4}{11}÷\frac{2}{11}) = 2×\frac{1}{5}×2 = \frac{4}{5}$　$\frac{2022×2023 - 1}{2022 + 2021×2023} = \frac{(2021 + 1)×2023 - 1}{2022 + 2021×2023} = \frac{2021×2023 + 2023 - 1}{2022 + 2021×2023} = \frac{2021×2023 + 2022}{2022 + 2021×2023} = 1$
$\frac{1}{21} + \frac{202}{2121} + \frac{50505}{212121} + \frac{13131313}{21212121} = \frac{1}{21} + \frac{2×101}{21×101} + \frac{5×10101}{21×10101} + \frac{13×1010101}{21×1010101} = \frac{1}{21} + \frac{2}{21} + \frac{5}{21} + \frac{13}{21} = 1$