三、算一算。
1. 直接写出得数。
$\frac{5}{7}+\frac{9}{7}=$
$\frac{8}{15}-\frac{1}{3}=$
$\frac{5}{6}+\frac{5}{12}=$
$3-\frac{13}{21}+\frac{7}{21}=$
$\frac{7}{16}-\frac{3}{16}=$
$\frac{4}{9}+1\frac{7}{9}=$
$\frac{5}{8}-\frac{7}{12}=$
$\frac{6}{7}+\frac{11}{12}+\frac{1}{7}=$

1. $2$ $\frac{1}{5}$ $\frac{5}{4}$ $2\frac{5}{7}$ $\frac{1}{4}$ $2\frac{2}{9}$ $\frac{1}{24}$ $1\frac{11}{12}$

2. 计算下面各题。(能简算的要简算)
$\frac{5}{8} + \frac{1}{4} - \frac{5}{12}$
$\frac{4}{7} + ( \frac{3}{7} - 0.45 )$
$\frac{6}{7} - \frac{13}{15} + \frac{8}{7} + \frac{3}{15}$
$\frac{7}{9} - ( \frac{13}{15} - \frac{2}{9} )$
$\frac{7}{18} + ( \frac{5}{12} - \frac{1}{3} )$
$\frac{3}{2} - ( 0.875 - \frac{1}{2} ) - \frac{1}{8}$

2.
$\begin{aligned}&\frac{5}{8}+\frac{1}{4}-\frac{5}{12}\\=&\frac{15}{24}+\frac{6}{24}-\frac{10}{24}\\=&\frac{11}{24}\end{aligned}$
$\begin{aligned}&\frac{4}{7}+(\frac{3}{7}-0.45)\\=&\frac{4}{7}+\frac{3}{7}-0.45\\=&1-0.45\\=&0.55\end{aligned}$
$\begin{aligned}&\frac{6}{7}-\frac{13}{15}+\frac{8}{7}+\frac{3}{15}\\=&(\frac{6}{7}+\frac{8}{7})-(\frac{13}{15}-\frac{3}{15})\\=&2-\frac{2}{3}\\=&1\frac{1}{3}\end{aligned}$
$\begin{aligned}&\frac{7}{9}-(\frac{13}{15}-\frac{2}{9})\\=&\frac{7}{9}+\frac{2}{9}-\frac{13}{15}\\=&1-\frac{13}{15}\\=&\frac{2}{15}\end{aligned}$
$\begin{aligned}&\frac{7}{18}+(\frac{5}{12}-\frac{1}{3})\\=&\frac{14}{36}+(\frac{15}{36}-\frac{12}{36})\\=&\frac{17}{36}\end{aligned}$
$\begin{aligned}&\frac{3}{2}-(0.875-\frac{1}{2})-\frac{1}{8}\\=&\frac{3}{2}-\frac{7}{8}+\frac{1}{2}-\frac{1}{8}\\=&(\frac{3}{2}+\frac{1}{2})-(\frac{7}{8}+\frac{1}{8})\\=&1\end{aligned}$

3. 解方程。
$x+\dfrac{5}{8}=\dfrac{5}{6}$
$\dfrac{7}{9}-x=\dfrac{1}{6}$
$x-\dfrac{2}{5}+\dfrac{3}{10}=\dfrac{11}{30}$

3.
$\begin{aligned}x+\frac{5}{8}&=\frac{5}{6}\\解:x+\frac{5}{8}-\frac{5}{8}&=\frac{5}{6}-\frac{5}{8}\\x&=\frac{5}{24}\end{aligned}$
$\begin{aligned}\frac{7}{9}-x&=\frac{1}{6}\\解:\frac{7}{9}-x+x&=\frac{1}{6}+x\\\frac{7}{9}-\frac{1}{6}&=\frac{1}{6}+x-\frac{1}{6}\\x&=\frac{11}{18}\end{aligned}$
$\begin{aligned}x-\frac{2}{5}+\frac{3}{10}&=\frac{11}{30}\\解:x-(\frac{2}{5}-\frac{3}{10})&=\frac{11}{30}\\x-\frac{1}{10}&=\frac{11}{30}\\x&=\frac{7}{15}\end{aligned}$

四、解决问题。
1.

1. 腰长是$\frac{3}{10}$dm时:$\frac{3}{10}+\frac{3}{10}=\frac{3}{5}$(dm)
$\frac{3}{5}＜\frac{4}{5}$,不符合三角形三边关系
所以三角形的腰长是$\frac{4}{5}$dm
$\frac{3}{10}+\frac{4}{5}+\frac{4}{5}=\frac{19}{10}$(dm)
答:这个三角形的周长是$\frac{19}{10}$dm。

2. 一节课45分钟，老师讲授时间占$\frac{1}{3}$，学生合作探究时间占$\frac{1}{6}$，讨论汇报时间占$\frac{1}{15}$，剩下的是学生独立完成作业时间，学生独立完成作业时间占几分之几？

2. $1-\frac{1}{3}-\frac{1}{6}-\frac{1}{15}=\frac{13}{30}$
答:学生独立完成作业时间占$\frac{13}{30}$。