1. 解方程。
(1)$\frac{2}{3}x + \frac{1}{2}x = 42$
(2)$\frac{1}{15}: x = \frac{1}{12}: 3$

解：
(1)
$\frac{4}{6}x + \frac{3}{6}x = 42$
$\frac{7}{6}x = 42$
$x = 42 × \frac{6}{7}$
$x = 36$
(2)
$\frac{1}{12}x = \frac{1}{15} × 3$
$\frac{1}{12}x = \frac{1}{5}$
$x = \frac{1}{5} ÷ \frac{1}{12}$
$x = \frac{12}{5}$

2. 用简便方法计算。
(1)$5.3 - 3\frac{3}{7} + 4.7 - 4\frac{4}{7}$
(2)$24×( \frac{1}{4} + \frac{5}{6} - \frac{7}{8} )$

(1)
$\begin{aligned}5.3 - 3\frac{3}{7} + 4.7 - 4\frac{4}{7}&=(5.3 + 4.7) - (3\frac{3}{7} + 4\frac{4}{7})\\&=10 - 8\\&=2\end{aligned}$
(2)
$\begin{aligned}24×(\frac{1}{4} + \frac{5}{6} - \frac{7}{8})&=24×\frac{1}{4} + 24×\frac{5}{6} - 24×\frac{7}{8}\\&=6 + 20 - 21\\&=5\end{aligned}$

3. 用递等式计算。
(1) $109 - 4.5 ÷ 0.05 + 91$
(2) $0.08 × (5.6 + 4.8) ÷ 1.3$
(3) $(0.75 - \dfrac{2}{3}) × (4 ÷ \dfrac{3}{4})$
(4) $\dfrac{8}{9} × [\dfrac{3}{4} - (\dfrac{7}{16} - \dfrac{1}{4})]$

(1)
$\begin{aligned}109 - 4.5 ÷ 0.05 + 91&=109 - 90 + 91\\&=19 + 91\\&=110\end{aligned}$
(2)
$\begin{aligned}0.08 × (5.6 + 4.8) ÷ 1.3&=0.08 × 10.4 ÷ 1.3\\&=0.832 ÷ 1.3\\&=0.64\end{aligned}$
(3)
$\begin{aligned}(0.75 - \dfrac{2}{3}) × (4 ÷ \dfrac{3}{4})&=(\dfrac{3}{4} - \dfrac{2}{3}) × (4 × \dfrac{4}{3})\\&=\dfrac{1}{12} × \dfrac{16}{3}\\&=\dfrac{4}{9}\end{aligned}$
(4)
$\begin{aligned}\dfrac{8}{9} × [\dfrac{3}{4} - (\dfrac{7}{16} - \dfrac{1}{4})]&=\dfrac{8}{9} × [\dfrac{3}{4} + \dfrac{1}{4} - \dfrac{7}{16}]\\&=\dfrac{8}{9} × [1 - \dfrac{7}{16}]\\&=\dfrac{8}{9} × \dfrac{9}{16}\\&=\dfrac{1}{2}\end{aligned}$

五、甲、乙两车行驶的路程与时间的关系如图：

1. 从图中可以看出，甲车行驶的路程与行驶的时间成()比例关系。
2. 如果甲、乙两车从A,B两地同时出发，相向而行，经过5小时相遇。则A,B两地相距多少千米？

1. 正
2. 甲车:270 ÷ 3 = 90(千米/时)
乙车:180 ÷ 3 = 60(千米/时)
(90 + 60) × 5 = 750(千米)
答:A,B两地相距750千米。