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1.(苏州真题)解方程。
$\frac {12}{5}x-\frac {7}{2}=\frac {13}{10}$ $6x+x=\frac {147}{10}$
$\frac {12}{5}x-\frac {7}{2}=\frac {13}{10}$ $6x+x=\frac {147}{10}$
答案
1. $\frac{12}{5}x - \frac{7}{2} = \frac{13}{10}$
解: $\frac{12}{5}x = \frac{13}{10} + \frac{7}{2}$
$\frac{12}{5}x = \frac{24}{5}$
$x = 2$
$6x + x = \frac{147}{10}$
解: $7x = \frac{147}{10}$
$x = \frac{21}{10}$
解: $\frac{12}{5}x = \frac{13}{10} + \frac{7}{2}$
$\frac{12}{5}x = \frac{24}{5}$
$x = 2$
$6x + x = \frac{147}{10}$
解: $7x = \frac{147}{10}$
$x = \frac{21}{10}$
2. 计算下面各题,能简算的要简算。
$\frac {16}{11}-(\frac {5}{11}-\frac {2}{9})$ $30×(\frac {1}{15}+\frac {2}{17})×17$
$\frac {4}{5}×\frac {5}{6}+\frac {2}{5}÷\frac {6}{5}$ $\frac {4}{5}÷[(\frac {5}{8}-\frac {1}{2})÷\frac {5}{8}]$
$\frac {16}{11}-(\frac {5}{11}-\frac {2}{9})$ $30×(\frac {1}{15}+\frac {2}{17})×17$
$\frac {4}{5}×\frac {5}{6}+\frac {2}{5}÷\frac {6}{5}$ $\frac {4}{5}÷[(\frac {5}{8}-\frac {1}{2})÷\frac {5}{8}]$
答案
2. $\frac{16}{11} - (\frac{5}{11} - \frac{2}{9})$
$= \frac{16}{11} - \frac{5}{11} + \frac{2}{9}$
$= 1 + \frac{2}{9}$
$= 1\frac{2}{9}$
$30×(\frac{1}{15} + \frac{2}{17})×17$
$= 30×\frac{1}{15}×17 + 30×\frac{2}{17}×17$
$= 34 + 60$
$= 94$
$\frac{4}{5}×\frac{5}{6} + \frac{2}{5}÷\frac{6}{5}$
$= \frac{4}{5}×\frac{5}{6} + \frac{2}{5}×\frac{5}{6}$
$= (\frac{4}{5} + \frac{2}{5})×\frac{5}{6}$
$= \frac{6}{5}×\frac{5}{6}$
$= 1$
$\frac{4}{5}÷[(\frac{5}{8} - \frac{1}{2})÷\frac{5}{8}]$
$= \frac{4}{5}÷[\frac{1}{8}÷\frac{5}{8}]$
$= \frac{4}{5}÷\frac{1}{5}$
$= 4$
$= \frac{16}{11} - \frac{5}{11} + \frac{2}{9}$
$= 1 + \frac{2}{9}$
$= 1\frac{2}{9}$
$30×(\frac{1}{15} + \frac{2}{17})×17$
$= 30×\frac{1}{15}×17 + 30×\frac{2}{17}×17$
$= 34 + 60$
$= 94$
$\frac{4}{5}×\frac{5}{6} + \frac{2}{5}÷\frac{6}{5}$
$= \frac{4}{5}×\frac{5}{6} + \frac{2}{5}×\frac{5}{6}$
$= (\frac{4}{5} + \frac{2}{5})×\frac{5}{6}$
$= \frac{6}{5}×\frac{5}{6}$
$= 1$
$\frac{4}{5}÷[(\frac{5}{8} - \frac{1}{2})÷\frac{5}{8}]$
$= \frac{4}{5}÷[\frac{1}{8}÷\frac{5}{8}]$
$= \frac{4}{5}÷\frac{1}{5}$
$= 4$
3. 选一选。
(1)(算法探究)计算$36×\frac {34}{35}$时,(
A.$(37-1)×\frac {34}{35}$ B.$(35+1)×\frac {34}{35}$
C.$36×(1-\frac {1}{35})$ D.$(35+1)×(1-\frac {1}{35})$
(2)如图,这个梯形的面积是(
A. 9 B.$\frac {9}{2}$ C.$\frac {9}{4}$ D.$\frac {11}{4}$
(1)(算法探究)计算$36×\frac {34}{35}$时,(
B
)最简便。A.$(37-1)×\frac {34}{35}$ B.$(35+1)×\frac {34}{35}$
C.$36×(1-\frac {1}{35})$ D.$(35+1)×(1-\frac {1}{35})$
(2)如图,这个梯形的面积是(
C
)$m^{2}$。A. 9 B.$\frac {9}{2}$ C.$\frac {9}{4}$ D.$\frac {11}{4}$
答案
3. (1) B (2) C
4.(易错题)瑞海修路队修一条路,前4小时一共修了$\frac {3}{10}$千米,后3小时每小时修$\frac {7}{20}$千米,平均每小时修多少千米?
答案
4. $(\frac{3}{10} + \frac{7}{20}×3)÷(4 + 3) = \frac{27}{140}$(千米)
易错分析:易错在把“后3小时每小时修$\frac{7}{20}$千米”看成“后3小时一共修了$\frac{7}{20}$千米”。
易错分析:易错在把“后3小时每小时修$\frac{7}{20}$千米”看成“后3小时一共修了$\frac{7}{20}$千米”。
5.
(1)两种香瓜子各买3包,一共多少千克?
(2)买64包A牌香瓜子需要多少元? 如果用这些钱买B牌香瓜子,那么可以买多少包?
(1)两种香瓜子各买3包,一共多少千克?
(2)买64包A牌香瓜子需要多少元? 如果用这些钱买B牌香瓜子,那么可以买多少包?
答案
5. (1) $(\frac{1}{4} + \frac{1}{5})×3 = \frac{27}{20}$(千克)
(2) $64×\frac{1}{4}×28 = 448$(元)
$448÷(32×\frac{1}{5}) = 70$(包)
(2) $64×\frac{1}{4}×28 = 448$(元)
$448÷(32×\frac{1}{5}) = 70$(包)
6.(1)(思维过程)明明在计算$3×(\frac {1}{5}+□)-\frac {1}{4}$时,把括号弄丢了,计算结果比正确结果小了$\frac {1}{3}$,□是(
(2)明明把$(\frac {4}{7}+a)×3$错当成$\frac {4}{7}+a×3$进行计算,这样算出的结果与正确结果相差(
$\frac{1}{6}$
)。(2)明明把$(\frac {4}{7}+a)×3$错当成$\frac {4}{7}+a×3$进行计算,这样算出的结果与正确结果相差(
$\frac{8}{7}$
)。答案
6. (1) $\frac{1}{6}$ 解析:将$3×(\frac{1}{5} + ■) - \frac{1}{4}$去括号,整理可得$3×\frac{1}{5} + 3×■ - \frac{1}{4}$;把括号弄丢后的算式为$3×\frac{1}{5} + ■ - \frac{1}{4}$,与正确的算式相比,少了2个
,即$■ = \frac{1}{3}÷2 = \frac{1}{6}$。(2) $\frac{8}{7}$ 解析:$(\frac{4}{7} + a)×3 = \frac{4}{7}×3 + a×3$,与$\frac{4}{7} + a×3$相比,$\frac{4}{7}×3$被当成了$\frac{4}{7}$,这样算出的结果与正确结果相差$\frac{4}{7}×3 - \frac{4}{7} = \frac{8}{7}$。
