题目内容
19.计算,能简算的要简算.| $\frac{10}{11}$×$\frac{1}{13}$×13×11 | 2004×$\frac{2005}{2006}$ | $\frac{13}{49}$×21+28×$\frac{13}{49}$ |
| $\frac{4}{5}$-$\frac{4}{5}$×$\frac{5}{6}$ | 72×$\frac{5}{71}$ | ($\frac{3}{4}$+$\frac{5}{8}$)×32 |
| $\frac{6}{25}$÷9+$\frac{6}{25}$×$\frac{8}{9}$ | $\frac{3}{20}$÷0.2×$\frac{2}{3}$ | $\frac{3}{4}$÷$\frac{7}{8}$÷$\frac{15}{14}$. |
分析 (1)运用乘法交换律和结合律简算;
(2)把2004看作2006-2,运用乘法分配律简算;
(3)(4)(6)(7)运用乘法分配律简算;
(5)把72看作71+1,运用乘法分配律简算;i
(8)把小数化为分数,把除法变为乘法,约分计算;
(9)把除法变为乘法,约分计算.
解答 解:(1)$\frac{10}{11}$×$\frac{1}{13}$×13×11
=($\frac{10}{11}$×11)×($\frac{1}{13}$×13)
=10×1
=10
(2)2004×$\frac{2005}{2006}$
=(2006-2)×$\frac{2005}{2006}$
=2006×$\frac{2005}{2006}$-2×$\frac{2005}{2006}$
=2005-1$\frac{1002}{1003}$
=2003$\frac{1}{1003}$
(3)$\frac{13}{49}$×21+28×$\frac{13}{49}$
=$\frac{13}{49}$×(21+28)
=$\frac{13}{49}$×49
=13
(4)$\frac{4}{5}$-$\frac{4}{5}$×$\frac{5}{6}$
=$\frac{4}{5}$×(1-$\frac{5}{6}$)
=$\frac{4}{5}$×$\frac{1}{6}$
=$\frac{2}{15}$
(5)72×$\frac{5}{71}$
=(71+1)×$\frac{5}{71}$
=71×$\frac{5}{71}$+71+$\frac{5}{71}$
=5+$\frac{5}{71}$
=5$\frac{5}{71}$
(6)($\frac{3}{4}$+$\frac{5}{8}$)×32
=$\frac{3}{4}$×32+$\frac{5}{8}$×32
=24+20
=44
(7)$\frac{6}{25}$÷9+$\frac{6}{25}$×$\frac{8}{9}$
=$\frac{6}{25}$×$\frac{1}{9}$+$\frac{6}{25}$×$\frac{8}{9}$
=$\frac{6}{25}$×($\frac{1}{9}$+$\frac{8}{9}$)
=$\frac{6}{25}$×1
=$\frac{6}{25}$
(8)$\frac{3}{20}$÷0.2×$\frac{2}{3}$
=$\frac{3}{20}$÷$\frac{1}{5}$×$\frac{2}{3}$
=$\frac{3}{20}$×5×$\frac{2}{3}$
=$\frac{1}{2}$
(9)$\frac{3}{4}$÷$\frac{7}{8}$÷$\frac{15}{14}$
=$\frac{3}{4}$×$\frac{8}{7}$×$\frac{14}{15}$
=$\frac{4}{5}$
点评 此题主要考查分数的四则混合运算的运算顺序和应用运算定律进行简便计算.
| A. | 6 | B. | 4 | C. | 5 |