解下列不等式组:
(1)$\begin{cases}2x + 1 ≤ x + 2,①\\2x - 1 ＞ \dfrac{1}{2}(x + 4);②\end{cases}$ (2)$\begin{cases}2(x + 1) ＞ x,①\\1 - 2x ≥ \dfrac{x + 7}{2};②\end{cases}$
(3)$\begin{cases}2x + 2 ＞ 3(x - 1),①\\\dfrac{1}{2}x - 1 ≤ 7 - \dfrac{3}{2}x;②\end{cases}$ (4)$\begin{cases}\dfrac{2x + 1}{3} - 2 ≥ 0,①\\5 - 2(x + 1) ＜ 0;②\end{cases}$
(5)$\begin{cases}2x + 5 ≤ 3(x + 2),①\\\dfrac{x - 1}{2} ＜ \dfrac{x}{3};②\end{cases}$ (6)$\begin{cases}\dfrac{3}{2}x + 5 ＞ 1 - x,①\\x - 1 ≤ \dfrac{3}{4}x - \dfrac{1}{8};②\end{cases}$
(7)$\begin{cases}x - \dfrac{3}{2}(x - 2) ≤ 5,①\\\dfrac{1 + 3x}{2} ＞ 2x - 1.②\end{cases}$
(1)$\begin{cases}2x + 1 ≤ x + 2,①\\2x - 1 ＞ \dfrac{1}{2}(x + 4);②\end{cases}$ (2)$\begin{cases}2(x + 1) ＞ x,①\\1 - 2x ≥ \dfrac{x + 7}{2};②\end{cases}$
(3)$\begin{cases}2x + 2 ＞ 3(x - 1),①\\\dfrac{1}{2}x - 1 ≤ 7 - \dfrac{3}{2}x;②\end{cases}$ (4)$\begin{cases}\dfrac{2x + 1}{3} - 2 ≥ 0,①\\5 - 2(x + 1) ＜ 0;②\end{cases}$
(5)$\begin{cases}2x + 5 ≤ 3(x + 2),①\\\dfrac{x - 1}{2} ＜ \dfrac{x}{3};②\end{cases}$ (6)$\begin{cases}\dfrac{3}{2}x + 5 ＞ 1 - x,①\\x - 1 ≤ \dfrac{3}{4}x - \dfrac{1}{8};②\end{cases}$
(7)$\begin{cases}x - \dfrac{3}{2}(x - 2) ≤ 5,①\\\dfrac{1 + 3x}{2} ＞ 2x - 1.②\end{cases}$
答案：解:(1)解不等式①,得$x≤ 1$;解不等式②,得$x＞2$.故原不等式组无解.
(2)解不等式①,得$x＞-2$;解不等式②,得$x≤ -1$.故原不等式组的解集为$-2＜x≤ -1$.
(3)解不等式①,得$x＜5$;解不等式②,得$x≤ 4$.故原不等式组的解集为$x≤ 4$.
(4)解不等式①,得$x≥ \frac{5}{2}$;解不等式②,得$x＞\frac{3}{2}$.故原不等式组的解集为$x≥ \frac{5}{2}$.
(5)解不等式①,得$x≥ -1$;解不等式②,得$x＜3$.故原不等式组的解集为$-1≤ x＜3$.
(6)解不等式①,得$x＞-\frac{8}{5}$;解不等式②,得$x≤ \frac{7}{2}$.故原不等式组的解集为$-\frac{8}{5}＜x≤ \frac{7}{2}$.
(7)解不等式①,得$x≥ -4$;解不等式②,得$x＜3$.故原不等式组的解集为$-4≤ x＜3$.
(2)解不等式①,得$x＞-2$;解不等式②,得$x≤ -1$.故原不等式组的解集为$-2＜x≤ -1$.
(3)解不等式①,得$x＜5$;解不等式②,得$x≤ 4$.故原不等式组的解集为$x≤ 4$.
(4)解不等式①,得$x≥ \frac{5}{2}$;解不等式②,得$x＞\frac{3}{2}$.故原不等式组的解集为$x≥ \frac{5}{2}$.
(5)解不等式①,得$x≥ -1$;解不等式②,得$x＜3$.故原不等式组的解集为$-1≤ x＜3$.
(6)解不等式①,得$x＞-\frac{8}{5}$;解不等式②,得$x≤ \frac{7}{2}$.故原不等式组的解集为$-\frac{8}{5}＜x≤ \frac{7}{2}$.
(7)解不等式①,得$x≥ -4$;解不等式②,得$x＜3$.故原不等式组的解集为$-4≤ x＜3$.