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算一算
$16× \dfrac{3}{8}=$
$\dfrac{5}{6}× 2=$
$\dfrac{5}{8}× \dfrac{4}{5}=$
$\dfrac{25}{36}× \dfrac{27}{40}=$
$\dfrac{18}{25}× \dfrac{5}{9}=$
$\dfrac{5}{27}× 120=$
$\dfrac{1}{39}× \dfrac{13}{32}=$
$\dfrac{5}{8}÷ \dfrac{3}{8}=$
$\dfrac{3}{12}÷ \dfrac{5}{12}=$
$16× \dfrac{3}{8}=$
$\dfrac{5}{6}× 2=$
$\dfrac{5}{8}× \dfrac{4}{5}=$
$\dfrac{25}{36}× \dfrac{27}{40}=$
$\dfrac{18}{25}× \dfrac{5}{9}=$
$\dfrac{5}{27}× 120=$
$\dfrac{1}{39}× \dfrac{13}{32}=$
$\dfrac{5}{8}÷ \dfrac{3}{8}=$
$\dfrac{3}{12}÷ \dfrac{5}{12}=$
答案
16×$\dfrac{3}{8}$ = (16÷8)×3 = 2×3 = 6
$\dfrac{5}{6}×2$ = $\dfrac{5×2}{6}$ = $\dfrac{10}{6}$ = $\dfrac{5}{3}$
$\dfrac{5}{8}×\dfrac{4}{5}$ = $\dfrac{5×4}{8×5}$ = $\dfrac{4}{8}$ = $\dfrac{1}{2}$
$\dfrac{25}{36}×\dfrac{27}{40}$ = $\dfrac{25×27}{36×40}$ = $\dfrac{5×3}{4×8}$ = $\dfrac{15}{32}$
$\dfrac{18}{25}×\dfrac{5}{9}$ = $\dfrac{18×5}{25×9}$ = $\dfrac{2×1}{5×1}$ = $\dfrac{2}{5}$
$\dfrac{5}{27}×120$ = $\dfrac{5×120}{27}$ = $\dfrac{5×40}{9}$ = $\dfrac{200}{9}$
$\dfrac{1}{39}×\dfrac{13}{32}$ = $\dfrac{1×13}{39×32}$ = $\dfrac{1×1}{3×32}$ = $\dfrac{1}{96}$
$\dfrac{5}{8}÷\dfrac{3}{8}$ = $\dfrac{5}{8}×\dfrac{8}{3}$ = $\dfrac{5}{3}$
$\dfrac{3}{12}÷\dfrac{5}{12}$ = $\dfrac{3}{12}×\dfrac{12}{5}$ = $\dfrac{3}{5}$
$\dfrac{5}{6}×2$ = $\dfrac{5×2}{6}$ = $\dfrac{10}{6}$ = $\dfrac{5}{3}$
$\dfrac{5}{8}×\dfrac{4}{5}$ = $\dfrac{5×4}{8×5}$ = $\dfrac{4}{8}$ = $\dfrac{1}{2}$
$\dfrac{25}{36}×\dfrac{27}{40}$ = $\dfrac{25×27}{36×40}$ = $\dfrac{5×3}{4×8}$ = $\dfrac{15}{32}$
$\dfrac{18}{25}×\dfrac{5}{9}$ = $\dfrac{18×5}{25×9}$ = $\dfrac{2×1}{5×1}$ = $\dfrac{2}{5}$
$\dfrac{5}{27}×120$ = $\dfrac{5×120}{27}$ = $\dfrac{5×40}{9}$ = $\dfrac{200}{9}$
$\dfrac{1}{39}×\dfrac{13}{32}$ = $\dfrac{1×13}{39×32}$ = $\dfrac{1×1}{3×32}$ = $\dfrac{1}{96}$
$\dfrac{5}{8}÷\dfrac{3}{8}$ = $\dfrac{5}{8}×\dfrac{8}{3}$ = $\dfrac{5}{3}$
$\dfrac{3}{12}÷\dfrac{5}{12}$ = $\dfrac{3}{12}×\dfrac{12}{5}$ = $\dfrac{3}{5}$
1. 甲数的$\frac{5}{9}$等于乙数的$\frac{5}{7}$,则甲数(
A.大于
B.等于
C.小于
A
)乙数。A.大于
B.等于
C.小于
答案
1.A
2. 将$\frac{4}{5}\ \mathrm{kg}$糖果平均分成4份,每份是1 kg的( $\quad\quad$ )。
A.$\frac{1}{5}$
B.$\frac{1}{20}$
C.$\frac{1}{4}$
A.$\frac{1}{5}$
B.$\frac{1}{20}$
C.$\frac{1}{4}$
答案
2.A
3. 若$\frac{5}{9} ÷ a ＞ \frac{5}{9} (a ≠ 0)$，那么$a$一定是(
A.真分数
B.假分数
C.无法确定
A
)。A.真分数
B.假分数
C.无法确定
答案
3.A
4. 下面算式中,得数最大的是(
A.$\frac{2007}{2008} ÷ \frac{5}{12}$
B.$\frac{2007}{2008} × \frac{5}{12}$
C.$\frac{2007}{2008} ÷ \frac{1}{2}$
A
)。A.$\frac{2007}{2008} ÷ \frac{5}{12}$
B.$\frac{2007}{2008} × \frac{5}{12}$
C.$\frac{2007}{2008} ÷ \frac{1}{2}$
答案
4.A
二、先将下面的等量关系表示出来,再列方程。
1. 水果店运来 $ x $ 箱苹果,每箱重 $ 10 \ \mathrm{kg} $,卖出 $ 75 \ \mathrm{kg} $,还剩下 $ 5 \ \mathrm{kg} $。
等量关系:
方程:
2. 一个长方形的宽为 $ x \ \mathrm{cm} $,长是宽的 $ 2 $ 倍,周长为 $ 162 \ \mathrm{cm} $。
等量关系:
方程:
1. 水果店运来 $ x $ 箱苹果,每箱重 $ 10 \ \mathrm{kg} $,卖出 $ 75 \ \mathrm{kg} $,还剩下 $ 5 \ \mathrm{kg} $。
等量关系:
$x$箱苹果的质量 - 卖出的质量 = 剩下的质量
方程:
$10x - 75 = 5$
2. 一个长方形的宽为 $ x \ \mathrm{cm} $,长是宽的 $ 2 $ 倍,周长为 $ 162 \ \mathrm{cm} $。
等量关系:
$2 × (长 + 宽) = 长方形的周长$
方程:
$2(x + 2x) = 162$
答案
1. 等量关系：$x$箱苹果的质量 - 卖出的质量 = 剩下的质量，方程：$10x - 75 = 5$
2. 等量关系：$2 × (长 + 宽) = 长方形的周长$，方程：$2(x + 2x) = 162$
2. 等量关系：$2 × (长 + 宽) = 长方形的周长$，方程：$2(x + 2x) = 162$
三、解方程。
$\frac{1}{3} + \frac{1}{6}x = \frac{5}{7}$
$5.4x + x = 128$
$0.3x = x - \frac{1}{10}$
$0.5x - \frac{1}{2} = \frac{3}{4}$
$\frac{1}{3} + \frac{1}{6}x = \frac{5}{7}$
$5.4x + x = 128$
$0.3x = x - \frac{1}{10}$
$0.5x - \frac{1}{2} = \frac{3}{4}$
答案
解：
1. 解方程 $\frac{1}{3} + \frac{1}{6}x = \frac{5}{7}$
移项得：$\frac{1}{6}x = \frac{5}{7} - \frac{1}{3}$
计算右侧：$\frac{1}{6}x = \frac{15}{21} - \frac{7}{21} = \frac{8}{21}$
两边同乘6：$x = \frac{8}{21} × 6 = \frac{16}{7}$
2. 解方程 $5.4x + x = 128$
合并同类项得：$6.4x = 128$
两边同除以6.4：$x = 128 ÷ 6.4 = 20$
3. 解方程 $0.3x = x - \frac{1}{10}$
移项得：$x - 0.3x = \frac{1}{10}$
合并同类项：$0.7x = 0.1$
两边同除以0.7：$x = \frac{0.1}{0.7} = \frac{1}{7}$
4. 解方程 $0.5x - \frac{1}{2} = \frac{3}{4}$
移项得：$0.5x = \frac{3}{4} + \frac{1}{2}$
计算右侧：$0.5x = \frac{3}{4} + \frac{2}{4} = \frac{5}{4}$
两边同乘2：$x = \frac{5}{4} × 2 = \frac{5}{2}$
1. 解方程 $\frac{1}{3} + \frac{1}{6}x = \frac{5}{7}$
移项得：$\frac{1}{6}x = \frac{5}{7} - \frac{1}{3}$
计算右侧：$\frac{1}{6}x = \frac{15}{21} - \frac{7}{21} = \frac{8}{21}$
两边同乘6：$x = \frac{8}{21} × 6 = \frac{16}{7}$
2. 解方程 $5.4x + x = 128$
合并同类项得：$6.4x = 128$
两边同除以6.4：$x = 128 ÷ 6.4 = 20$
3. 解方程 $0.3x = x - \frac{1}{10}$
移项得：$x - 0.3x = \frac{1}{10}$
合并同类项：$0.7x = 0.1$
两边同除以0.7：$x = \frac{0.1}{0.7} = \frac{1}{7}$
4. 解方程 $0.5x - \frac{1}{2} = \frac{3}{4}$
移项得：$0.5x = \frac{3}{4} + \frac{1}{2}$
计算右侧：$0.5x = \frac{3}{4} + \frac{2}{4} = \frac{5}{4}$
两边同乘2：$x = \frac{5}{4} × 2 = \frac{5}{2}$
