1. 一题多解 解方程：$\frac{3}{2}[\frac{2}{3}(\frac{1}{2}x - \frac{1}{4}) - 4] = \frac{3}{2}x + 1.$

解法一:去中括号,得$(\dfrac{1}{2}x-\dfrac{1}{4})-6=\dfrac{3}{2}x+1$. 去括号,得$\dfrac{1}{2}x-\dfrac{1}{4}-6=\dfrac{3}{2}x+1$. 去分母,得$2x-1-24=6x+4$. 移项,得$2x-6x=4+24+1$. 合并同类项,得$-4x=29$. 系数化为1,得$x=-\dfrac{29}{4}$.
解法二:去小括号,得$\dfrac{3}{2}(\dfrac{1}{3}x-\dfrac{1}{6}-4)=\dfrac{3}{2}x+1$. 去括号,得$\dfrac{1}{2}x-\dfrac{1}{4}-6=\dfrac{3}{2}x+1$. 移项,得$\dfrac{1}{2}x-\dfrac{3}{2}x=1+\dfrac{1}{4}+6$. 合并同类项,得$-x=\dfrac{29}{4}$. 系数化为1,得$x=-\dfrac{29}{4}$.

2. 解方程：
(1) $\frac{0.1y - 0.2}{0.02} - \frac{y + 1}{0.5} = 3.$
(2) $\frac{4 - 6x}{0.03} - 6.5 = \frac{0.02 - 2x}{0.02} - 7.5.$

(1) 将分母化为整数,得$\dfrac{10y-20}{2}-\dfrac{10y+10}{5}=3$. 约分,得$5y-10-2y-2=3$. 移项、合并同类项,得$3y=15$. 系数化为1,得$y=5$.
(2) 原方程可化为$\dfrac{400-600x}{3}=\dfrac{2-200x}{2}-1$. 去分母,得$400-600x=3-300x-3$. 移项、合并同类项,得$-300x=-400$,解得$x=\dfrac{4}{3}$.

3. 解方程：$5(2x+3)-\frac{3}{4}(x-2)=2(x-2)-\frac{1}{2}(2x+3).$

移项,得$5(2x+3)+\dfrac{1}{2}(2x+3)=2(x-2)+\dfrac{3}{4}(x-2)$. 合并同类项,得$\dfrac{11}{2}(2x+3)=\dfrac{11}{4}(x-2)$. 去分母、去括号,得$4x+6=x-2$. 移项、合并同类项,得$3x=-8$. 系数化为1,得$x=-\dfrac{8}{3}$.

4. 解方程：
(1) $\frac{x}{3}+\frac{x-2}{5}=\frac{24}{7}-\frac{6-3x}{15}$.
(2) $\frac{1}{2}(y+1)+\frac{1}{3}(y+2)=3-\frac{1}{4}(y+3)$.

(1) 原方程可化为$\dfrac{x}{3}+\dfrac{x-2}{5}=\dfrac{24}{7}+\dfrac{x-2}{5}$,即$\dfrac{x}{3}=\dfrac{24}{7}$,所以$x=\dfrac{72}{7}$.
(2) 原方程可化为$(\dfrac{y+1}{2}-1)+(\dfrac{y+2}{3}-1)+(\dfrac{y+3}{4}-1)=0$,即$\dfrac{y-1}{2}+\dfrac{y-1}{3}+\dfrac{y-1}{4}=0$,整理,得$(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4})(y-1)=0$,所以$y-1=0$,解得$y=1$.