《2025年同步练习册人民教育出版社七年级数学上册人教版山东专版》

解下列方程：
(1)$\dfrac{3x - 1}{4} - 1 = \dfrac{x + 4}{2}$；
(2)$x - \dfrac{1}{2}\left[x - \dfrac{1}{2}\left(x - \dfrac{1}{2}\right)\right] = 2$.
解(1)$\dfrac{3x - 1}{4} - 1 = \dfrac{x + 4}{2}$,
去分母(方程两边乘 4),
得$3x - 1 - 4 = 2(x + 4)$.
去括号,得$3x - 1 - 4 = 2x + 8$.
移项,得$3x - 2x = 8 + 1 + 4$.
合并同类项,得$x = 13$.
(2)$x - \dfrac{1}{2}\left[x - \dfrac{1}{2}\left(x - \dfrac{1}{2}\right)\right] = 2$.
去括号,得$x - \dfrac{1}{2}x + \dfrac{1}{4}x - \dfrac{1}{8} = 2$.
去分母(方程两边乘 8),得$8x - 4x + 2x - 1 = 16$.
移项,得$8x - 4x + 2x = 16 + 1$.
合并同类项,得$6x = 17$.
系数化为 1,得$x = \dfrac{17}{6}$.

答案：
(1)$\dfrac{3x - 1}{4} - 1 = \dfrac{x + 4}{2}$
去分母，得$3x - 1 - 4 = 2(x + 4)$
去括号，得$3x - 1 - 4 = 2x + 8$
移项，得$3x - 2x = 8 + 1 + 4$
合并同类项，得$x = 13$
(2)$x - \dfrac{1}{2}\left[x - \dfrac{1}{2}\left(x - \dfrac{1}{2}\right)\right] = 2$
去括号，得$x - \dfrac{1}{2}x + \dfrac{1}{4}x - \dfrac{1}{8} = 2$
去分母，得$8x - 4x + 2x - 1 = 16$
移项，得$8x - 4x + 2x = 16 + 1$
合并同类项，得$6x = 17$
系数化为 1，得$x = \dfrac{17}{6}$