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1. 在横线上填上适当的式子，把下列运算过程补充完整：
$1 - [ (-\frac{1}{3})^2 + (-\frac{2}{3})^3 ] × (-3)^4$
$= 1 - [ \_\_\_\_\_\_ + \_\_\_\_\_\_ ] × \_\_\_\_\_\_$
$= 1 - [ \_\_\_\_\_\_ × \_\_\_\_\_\_ - \_\_\_\_\_\_ × \_\_\_\_\_\_ ]$
$= 1 + \_\_\_\_\_\_ = \_\_\_\_\_\_.$
$[ 3 - ( \frac{2}{3} - 0.5 × \frac{1}{3} ) ] × [ \frac{1}{2} - (-2)^2 ]$
$= [ 3 - ( \frac{2}{3} - \_\_\_\_\_\_ ) ] × [ \frac{1}{2} - \_\_\_\_\_\_ ]$
$= [ 3 - \_\_\_\_\_\_ ] × ( \_\_\_\_\_\_ )$
$= \_\_\_\_\_\_ × ( \_\_\_\_\_\_ ) = -\frac{35}{4} = -8\frac{3}{4}.$
$3\frac{3}{4} × ( -\frac{2}{5} )^2 + \frac{1}{2} × ( -\frac{6}{7} ) ÷ ( \frac{1}{2} - 2 )$
$= \frac{15}{4} × \_\_\_\_\_\_ + ( \_\_\_\_\_\_ ) ÷ ( \_\_\_\_\_\_ )$
$= \_\_\_\_\_\_ + ( \_\_\_\_\_\_ ) × ( \_\_\_\_\_\_ )$
$= \_\_\_\_\_\_ + \_\_\_\_\_\_ = \frac{31}{35}.$
$[ 1\frac{3}{5} × ( \frac{4}{9} - 1 ) ]^2 ÷ [ \frac{5}{6} × ( -\frac{2}{5} ) ]^3$
$= [ \frac{8}{5} × \_\_\_\_\_\_ ]^2 ÷ [ \_\_\_\_\_\_ ]^3$
$= ( \_\_\_\_\_\_ )^2 ÷ ( \_\_\_\_\_\_ )$
$= \_\_\_\_\_\_ × \_\_\_\_\_\_ = -\frac{64}{3} = -21\frac{1}{3}.$
$1 - [ (-\frac{1}{3})^2 + (-\frac{2}{3})^3 ] × (-3)^4$
$= 1 - [ \_\_\_\_\_\_ + \_\_\_\_\_\_ ] × \_\_\_\_\_\_$
$= 1 - [ \_\_\_\_\_\_ × \_\_\_\_\_\_ - \_\_\_\_\_\_ × \_\_\_\_\_\_ ]$
$= 1 + \_\_\_\_\_\_ = \_\_\_\_\_\_.$
$[ 3 - ( \frac{2}{3} - 0.5 × \frac{1}{3} ) ] × [ \frac{1}{2} - (-2)^2 ]$
$= [ 3 - ( \frac{2}{3} - \_\_\_\_\_\_ ) ] × [ \frac{1}{2} - \_\_\_\_\_\_ ]$
$= [ 3 - \_\_\_\_\_\_ ] × ( \_\_\_\_\_\_ )$
$= \_\_\_\_\_\_ × ( \_\_\_\_\_\_ ) = -\frac{35}{4} = -8\frac{3}{4}.$
$3\frac{3}{4} × ( -\frac{2}{5} )^2 + \frac{1}{2} × ( -\frac{6}{7} ) ÷ ( \frac{1}{2} - 2 )$
$= \frac{15}{4} × \_\_\_\_\_\_ + ( \_\_\_\_\_\_ ) ÷ ( \_\_\_\_\_\_ )$
$= \_\_\_\_\_\_ + ( \_\_\_\_\_\_ ) × ( \_\_\_\_\_\_ )$
$= \_\_\_\_\_\_ + \_\_\_\_\_\_ = \frac{31}{35}.$
$[ 1\frac{3}{5} × ( \frac{4}{9} - 1 ) ]^2 ÷ [ \frac{5}{6} × ( -\frac{2}{5} ) ]^3$
$= [ \frac{8}{5} × \_\_\_\_\_\_ ]^2 ÷ [ \_\_\_\_\_\_ ]^3$
$= ( \_\_\_\_\_\_ )^2 ÷ ( \_\_\_\_\_\_ )$
$= \_\_\_\_\_\_ × \_\_\_\_\_\_ = -\frac{64}{3} = -21\frac{1}{3}.$
答案
$\frac{1}{9}, (-\frac{8}{27}), 81, \frac{1}{9}×81, \frac{8}{27}×81, 15, 16; \frac{1}{6}, 4, \frac{1}{2}, -\frac{7}{2}, \frac{5}{2}, -\frac{7}{2}; \frac{4}{25}, -\frac{3}{7}, -\frac{3}{2}, \frac{3}{5}, -\frac{3}{7}, -\frac{2}{3}, \frac{3}{5}, \frac{2}{7}; (-\frac{5}{9}), -\frac{1}{3}, -\frac{8}{9}, -\frac{1}{27}, \frac{64}{81}, (-27).$
2.计算$(-4)×\frac{1}{7}×(-0.25)×21$，为了使过程简便，运用的运算律是……【 】
A.乘法对加法的分配律
B.乘法结合律和乘法对加法的分配律
C.乘法交换律和乘法结合律
D.乘法交换律和乘法对加法的分配律
A.乘法对加法的分配律
B.乘法结合律和乘法对加法的分配律
C.乘法交换律和乘法结合律
D.乘法交换律和乘法对加法的分配律
答案
C
3. 下列计算：①$\frac{1}{3} - \frac{1}{3}×2 = 0×2 = 0$；
②$6÷(\frac{2}{3} - \frac{3}{2}) = 6÷\frac{2}{3} - 6÷\frac{3}{2} = 9 - 4 = 5$；
③$(-1)÷(-5)×(-\frac{1}{5}) = (-1)÷1 = -1$。
计算过程错误的有 ……………………【 】
A.①②
B.②③
C.①③
D.①②③
②$6÷(\frac{2}{3} - \frac{3}{2}) = 6÷\frac{2}{3} - 6÷\frac{3}{2} = 9 - 4 = 5$；
③$(-1)÷(-5)×(-\frac{1}{5}) = (-1)÷1 = -1$。
计算过程错误的有 ……………………【 】
A.①②
B.②③
C.①③
D.①②③
答案
D
4.下列计算错误的是 …………【 】
A.$1 + 6 × (-\frac{1}{6}) ÷ (-6) = \frac{7}{6}$
B.$(-6) ÷ (-4) ÷ (+1\frac{1}{4}) = \frac{6}{5}$
C.$-\frac{1}{30} ÷ (\frac{1}{3} + \frac{1}{6} - \frac{2}{5}) = -3$
D.$(-13\frac{1}{3}) ÷ 5 - 1\frac{2}{3} ÷ 5 + 13 × \frac{1}{4} = \frac{1}{4}$
A.$1 + 6 × (-\frac{1}{6}) ÷ (-6) = \frac{7}{6}$
B.$(-6) ÷ (-4) ÷ (+1\frac{1}{4}) = \frac{6}{5}$
C.$-\frac{1}{30} ÷ (\frac{1}{3} + \frac{1}{6} - \frac{2}{5}) = -3$
D.$(-13\frac{1}{3}) ÷ 5 - 1\frac{2}{3} ÷ 5 + 13 × \frac{1}{4} = \frac{1}{4}$
答案
C
5. 下列计算正确的是 …………【 】
A.$-24× \frac{3}{7} -3× (-\frac{3}{7}) = -12$
B.$(\frac{1}{2}-\frac{1}{3})÷(\frac{1}{3}-\frac{1}{4})×(\frac{1}{4}-\frac{1}{5}) = \frac{3}{10}$
C.$(-\frac{2}{3}+\frac{4}{5})÷(-\frac{1}{15}) = -2$
D.$-\frac{3}{4}÷(+\frac{4}{3})×(-\frac{8}{27}) = -\frac{1}{6}$
A.$-24× \frac{3}{7} -3× (-\frac{3}{7}) = -12$
B.$(\frac{1}{2}-\frac{1}{3})÷(\frac{1}{3}-\frac{1}{4})×(\frac{1}{4}-\frac{1}{5}) = \frac{3}{10}$
C.$(-\frac{2}{3}+\frac{4}{5})÷(-\frac{1}{15}) = -2$
D.$-\frac{3}{4}÷(+\frac{4}{3})×(-\frac{8}{27}) = -\frac{1}{6}$
答案
C
