$\frac{7}{8}-\frac{1}{4}+\frac{1}{8}=$
$\frac{1}{2}+2+\frac{1}{2}=$
$\frac{2}{3}+\frac{5}{6}-\frac{1}{6}=$
$\frac{1}{14}+\frac{5}{14}-\frac{3}{14}=$
$1-\frac{7}{12}-\frac{5}{12}=$
$\frac{4}{9}+\frac{1}{3}+1=$
$\frac{5}{7}-\frac{1}{7}+\frac{2}{7}=$
$\frac{11}{12}-\frac{5}{12}+\frac{1}{12}=$
$1-\frac{7}{15}+\frac{4}{15}=$
$\frac{1}{2}+\frac{7}{10}-\frac{1}{5}=$
$\frac{1}{2}-\frac{1}{3}+\frac{1}{4}=$
$\frac{4}{5}-\frac{3}{10}+\frac{1}{10}=$
$\frac{7}{8}-\frac{1}{2}-\frac{1}{8}=$
$1-\frac{7}{10}+\frac{1}{5}=$
$\frac{5}{9}+\frac{4}{9}-\frac{3}{8}=$
$\frac{5}{8}-\frac{1}{6}+\frac{3}{8}=$
$1-\frac{6}{13}+\frac{7}{13}=$
$\frac{2}{5}+\frac{1}{3}-\frac{1}{15}=$

$\frac{3}{4},3,\frac{4}{3},\frac{3}{14},0,1\frac{7}{9},\frac{6}{7},\frac{7}{12},\frac{4}{5},1,\frac{5}{12},\frac{3}{5},\frac{1}{4},\frac{1}{2},\frac{5}{8},\frac{5}{6},1\frac{1}{13},\frac{2}{3}$

1. $\frac{7}{8}-\frac{1}{4}+\frac{1}{8}=(\frac{7}{8}+\frac{1}{8})-\frac{1}{4}=1-\frac{1}{4}=\frac{3}{4}$
2. $\frac{1}{2}+2+\frac{1}{2}=(\frac{1}{2}+\frac{1}{2})+2=1+2=3$
3. $\frac{2}{3}+\frac{5}{6}-\frac{1}{6}=\frac{2}{3}+(\frac{5}{6}-\frac{1}{6})=\frac{2}{3}+\frac{4}{6}=\frac{2}{3}+\frac{2}{3}=\frac{4}{3}$
4. $\frac{1}{14}+\frac{5}{14}-\frac{3}{14}=\frac{1+5-3}{14}=\frac{3}{14}$
5. $1-\frac{7}{12}-\frac{5}{12}=1-(\frac{7}{12}+\frac{5}{12})=1-1=0$
6. $\frac{4}{9}+\frac{1}{3}+1=\frac{4}{9}+\frac{3}{9}+1=\frac{7}{9}+1=1\frac{7}{9}$
7. $\frac{5}{7}-\frac{1}{7}+\frac{2}{7}=\frac{5-1+2}{7}=\frac{6}{7}$
8. $\frac{11}{12}-\frac{5}{12}+\frac{1}{12}=\frac{11-5+1}{12}=\frac{7}{12}$
9. $1-\frac{7}{15}+\frac{4}{15}=1-(\frac{7}{15}-\frac{4}{15})=1-\frac{3}{15}=1-\frac{1}{5}=\frac{4}{5}$
10. $\frac{1}{2}+\frac{7}{10}-\frac{1}{5}=\frac{5}{10}+\frac{7}{10}-\frac{2}{10}=\frac{10}{10}=1$
11. $\frac{1}{2}-\frac{1}{3}+\frac{1}{4}=\frac{6}{12}-\frac{4}{12}+\frac{3}{12}=\frac{5}{12}$
12. $\frac{4}{5}-\frac{3}{10}+\frac{1}{10}=\frac{8}{10}-\frac{3}{10}+\frac{1}{10}=\frac{6}{10}=\frac{3}{5}$
13. $\frac{7}{8}-\frac{1}{2}-\frac{1}{8}=(\frac{7}{8}-\frac{1}{8})-\frac{1}{2}=\frac{6}{8}-\frac{4}{8}=\frac{2}{8}=\frac{1}{4}$
14. $1-\frac{7}{10}+\frac{1}{5}=\frac{10}{10}-\frac{7}{10}+\frac{2}{10}=\frac{5}{10}=\frac{1}{2}$
15. $\frac{5}{9}+\frac{4}{9}-\frac{3}{8}=1-\frac{3}{8}=\frac{5}{8}$
16. $\frac{5}{8}-\frac{1}{6}+\frac{3}{8}=(\frac{5}{8}+\frac{3}{8})-\frac{1}{6}=1-\frac{1}{6}=\frac{5}{6}$
17. $1-\frac{6}{13}+\frac{7}{13}=1+(\frac{7}{13}-\frac{6}{13})=1+\frac{1}{13}=1\frac{1}{13}$
18. $\frac{2}{5}+\frac{1}{3}-\frac{1}{15}=\frac{6}{15}+\frac{5}{15}-\frac{1}{15}=\frac{10}{15}=\frac{2}{3}$

连一连
$\frac{1}{2}-(\frac{1}{4}-\frac{1}{6})$
$\frac{1}{2}+\frac{1}{4}+\frac{1}{6}$
$\frac{1}{2}+(\frac{1}{4}-\frac{1}{6})$
$\frac{1}{2}-\frac{1}{4}-\frac{1}{6}$
$\frac{1}{2}+(\frac{1}{4}+\frac{1}{6})$
$\frac{1}{2}+\frac{1}{4}-\frac{1}{6}$
$\frac{1}{2}-(\frac{1}{4}+\frac{1}{6})$
$\frac{1}{2}-\frac{1}{4}+\frac{1}{6}$

$\frac{1}{2}-(\frac{1}{4}-\frac{1}{6})$连$\frac{1}{2}-\frac{1}{4}+\frac{1}{6}$；$\frac{1}{2}+\frac{1}{4}+\frac{1}{6}$连$\frac{1}{2}+(\frac{1}{4}+\frac{1}{6})$；$\frac{1}{2}+(\frac{1}{4}-\frac{1}{6})$连$\frac{1}{2}+\frac{1}{4}-\frac{1}{6}$；$\frac{1}{2}-\frac{1}{4}-\frac{1}{6}$连$\frac{1}{2}-(\frac{1}{4}+\frac{1}{6})$

根据分数加减混合运算去括号法则：括号前是“+”，去括号后符号不变；括号前是“-”，去括号后括号内符号变反。
$\frac{1}{2}-(\frac{1}{4}-\frac{1}{6})=\frac{1}{2}-\frac{1}{4}+\frac{1}{6}$
$\frac{1}{2}+\frac{1}{4}+\frac{1}{6}=\frac{1}{2}+(\frac{1}{4}+\frac{1}{6})$
$\frac{1}{2}+(\frac{1}{4}-\frac{1}{6})=\frac{1}{2}+\frac{1}{4}-\frac{1}{6}$
$\frac{1}{2}-\frac{1}{4}-\frac{1}{6}=\frac{1}{2}-(\frac{1}{4}+\frac{1}{6})$