一、我会填。
$\frac{1}{2} + \frac{1}{3} = (\quad) + (\quad) = (\quad)$
$\frac{1}{3} + \frac{1}{9} = (\quad) + (\quad) = (\quad)$

第一组依次填$\frac{3}{6}$、$\frac{2}{6}$、$\frac{5}{6}$；第二组依次填$\frac{3}{9}$、$\frac{1}{9}$、$\frac{4}{9}$

异分母分数相加时，先通分转化为同分母分数，再将分子相加，分母不变。第一题：$\frac{1}{2}+\frac{1}{3}=\frac{3}{6}+\frac{2}{6}=\frac{5}{6}$；第二题：$\frac{1}{3}+\frac{1}{9}=\frac{3}{9}+\frac{1}{9}=\frac{4}{9}$。

1. 直接写出得数。
$\frac{1}{3}+\frac{1}{4}=$
$\frac{10}{7}-1=$
$\frac{1}{8}+\frac{3}{4}=$
$\frac{5}{6}-\frac{1}{2}=$
$\frac{4}{5}-\frac{1}{3}=$
$\frac{1}{6}+\frac{1}{3}=$
$\frac{1}{4}-\frac{1}{5}=$
$\frac{4}{9}+\frac{5}{9}=$

$\frac{7}{12}$；$\frac{3}{7}$；$\frac{7}{8}$；$\frac{1}{3}$；$\frac{7}{15}$；$\frac{1}{2}$；$\frac{1}{20}$；1

异分母分数相加减，先通分转化为同分母分数，再按同分母分数加减法法则计算（分子相加减，分母不变），结果能约分的要约分。逐个计算如下：
$\frac{1}{3}+\frac{1}{4}=\frac{7}{12}$；
$\frac{10}{7}-1=\frac{3}{7}$；
$\frac{1}{8}+\frac{3}{4}=\frac{7}{8}$；
$\frac{5}{6}-\frac{1}{2}=\frac{1}{3}$；
$\frac{4}{5}-\frac{1}{3}=\frac{7}{15}$；
$\frac{1}{6}+\frac{1}{3}=\frac{1}{2}$；
$\frac{1}{4}-\frac{1}{5}=\frac{1}{20}$；
$\frac{4}{9}+\frac{5}{9}=1$

2. 解方程。
$x+\dfrac{1}{8}=\dfrac{3}{4}$
$x-\dfrac{2}{3}=\dfrac{1}{6}$
$x+\dfrac{1}{13}=\dfrac{5}{4}$
$x-\dfrac{5}{6}=\dfrac{3}{8}$
$x÷0.9=81$
$2.4x-x=12.6$

$x=\dfrac{5}{8}$；$x=\dfrac{5}{6}$；$x=\dfrac{61}{52}$；$x=\dfrac{29}{24}$；$x=72.9$；$x=9$

1. 解方程$x+\dfrac{1}{8}=\dfrac{3}{4}$：根据等式性质，两边同时减$\dfrac{1}{8}$，得$x=\dfrac{3}{4}-\dfrac{1}{8}=\dfrac{6}{8}-\dfrac{1}{8}=\dfrac{5}{8}$；2. 解方程$x-\dfrac{2}{3}=\dfrac{1}{6}$：两边同时加$\dfrac{2}{3}$，得$x=\dfrac{1}{6}+\dfrac{2}{3}=\dfrac{1}{6}+\dfrac{4}{6}=\dfrac{5}{6}$；3. 解方程$x+\dfrac{1}{13}=\dfrac{5}{4}$：两边同时减$\dfrac{1}{13}$，得$x=\dfrac{5}{4}-\dfrac{1}{13}=\dfrac{65}{52}-\dfrac{4}{52}=\dfrac{61}{52}$；4. 解方程$x-\dfrac{5}{6}=\dfrac{3}{8}$：两边同时加$\dfrac{5}{6}$，得$x=\dfrac{3}{8}+\dfrac{5}{6}=\dfrac{9}{24}+\dfrac{20}{24}=\dfrac{29}{24}$；5. 解方程$x÷0.9=81$：两边同时乘0.9，得$x=81×0.9=72.9$；6. 解方程$2.4x -x=12.6$：先化简左边得$1.4x=12.6$，两边同时除以1.4，得$x=12.6÷1.4=9$。

3. 灵活计算。
$1-(\dfrac{5}{6}+\dfrac{1}{12})$
$\dfrac{2}{3}-\dfrac{1}{8}-\dfrac{1}{6}$
$\dfrac{19}{16}-(\dfrac{3}{16}+\dfrac{7}{8})$
$\dfrac{8}{13}+\dfrac{15}{17}+\dfrac{18}{13}+\dfrac{2}{17}$
$\dfrac{14}{15}-\dfrac{1}{3}-\dfrac{3}{5}$
$\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{1}{6}$
$\dfrac{4}{5}-\dfrac{3}{10}-\dfrac{1}{5}$
$\dfrac{2}{3}-(\dfrac{5}{12}-\dfrac{1}{4})$
$\dfrac{11}{12}-\dfrac{3}{10}+\dfrac{1}{12}-\dfrac{7}{10}$
$\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{5}{6}$

$\dfrac{1}{12}$；$\dfrac{3}{8}$；$\dfrac{1}{8}$；$3$；$0$；$\dfrac{19}{12}$；$\dfrac{3}{10}$；$\dfrac{1}{2}$；$0$；$\dfrac{23}{24}$

本题为分数加减混合运算，可利用加法交换律、结合律及去括号法则简便计算，异分母分数相加减先通分，再按同分母分数法则计算，结果化为最简分数。
1. $1-(\dfrac{5}{6}+\dfrac{1}{12})$：先算括号内，通分得$\dfrac{10}{12}+\dfrac{1}{12}=\dfrac{11}{12}$，再算$1-\dfrac{11}{12}=\dfrac{1}{12}$；
2. $\dfrac{2}{3}-\dfrac{1}{8}-\dfrac{1}{6}$：通分（分母24）得$\dfrac{16}{24}-\dfrac{3}{24}-\dfrac{4}{24}=\dfrac{9}{24}=\dfrac{3}{8}$；
3. $\dfrac{19}{16}-(\dfrac{3}{16}+\dfrac{7}{8})$：去括号得$\dfrac{19}{16}-\dfrac{3}{16}-\dfrac{14}{16}=\dfrac{2}{16}=\dfrac{1}{8}$；
4. $\dfrac{8}{13}+\dfrac{15}{17}+\dfrac{18}{13}+\dfrac{2}{17}$：加法交换律、结合律得$(\dfrac{8}{13}+\dfrac{18}{13})+(\dfrac{15}{17}+\dfrac{2}{17})=2+1=3$；
5. $\dfrac{14}{15}-\dfrac{1}{3}-\dfrac{3}{5}$：通分（分母15）得$\dfrac{14}{15}-\dfrac{5}{15}-\dfrac{9}{15}=0$；
6. $\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{1}{6}$：通分（分母12）得$\dfrac{8}{12}+\dfrac{9}{12}+\dfrac{2}{12}=\dfrac{19}{12}$；
7. $\dfrac{4}{5}-\dfrac{3}{10}-\dfrac{1}{5}$：交换位置得$\dfrac{4}{5}-\dfrac{1}{5}-\dfrac{3}{10}=\dfrac{3}{5}-\dfrac{3}{10}=\dfrac{3}{10}$；
8. $\dfrac{2}{3}-(\dfrac{5}{12}-\dfrac{1}{4})$：括号内计算得$\dfrac{5}{12}-\dfrac{3}{12}=\dfrac{1}{6}$，再算$\dfrac{2}{3}-\dfrac{1}{6}=\dfrac{1}{2}$；
9. $\dfrac{11}{12}-\dfrac{3}{10}+\dfrac{1}{12}-\dfrac{7}{10}$：加法交换律、结合律得$(\dfrac{11}{12}+\dfrac{1}{12})-(\dfrac{3}{10}+\dfrac{7}{10})=1-1=0$；
10. $\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{5}{6}$：通分（分母24）得$\dfrac{6}{24}-\dfrac{3}{24}+\dfrac{20}{24}=\dfrac{23}{24}$。