《2025年名校课堂七年级数学上册人教版陕西专版》

1. 解下列方程：
(1) $ 2y + 3 = 11 - 6y $.
(2) $ 5(x - 6) = -4x - 3 $.
(3) $ \frac{2x - 1}{-3} = \frac{5x + 4}{6} $.
(4) $ \frac{1}{6}(3x - 6) = \frac{2}{5}x - 3 $.
(5) $ 3(2x - 3) + 6 = 4(2x - 3) + 7 $.
(6) $ \frac{2}{5}x - 8 = \frac{1}{4} - \frac{1}{5}x $.
(7) $ \frac{4x - 1}{6} = 1 - \frac{3x - 2}{3} $.
(8) $ 2[x - 4(x - 1)] - 8 = -3 $.
(9) $ 3 - \frac{x - 1}{2} = x - \frac{11 + x}{4} $.
(10) $ \frac{x - 3}{0.2} - \frac{x + 2}{0.5} = 3 $.
(11) $ \frac{5 - x}{3} = \frac{x - 2}{2} - \frac{x + 1}{4} $.
(12) $ \frac{x + 1}{2} - \frac{x + 4}{6} = 1 + \frac{2}{3}x $.

答案： 1.解：
(1)移项，得2y + 6y = 11 - 3.合并同类项，得8y = 8.系数化为1，得y = 1.
(2)去括号，得5x - 30 = -4x - 3.移项，得5x + 4x = 30 - 3.合并同类项，得9x = 27.系数化为1，得x = 3.
(3)去分母，得-2(2x - 1) = 5x + 4.去括号，得-4x + 2 = 5x + 4.移项、合并同类项，得-9x = 2.系数化为1，得$x = -\frac{2}{9}. (4)$去分母，得5(3x - 6) = 12x - 90.去括号，得15x - 30 = 12x - 90.
移项、合并同类项，得3x = -60.系数化为1，得x = -20.
(5)(方法一)去括号，得6x - 9 + 6 = 8x - 12 + 7.移项，得6x - 8x = -12 + 7 + 9 - 6.合并同类项，得-2x = -2.系数化为1，得x = 1.(方法二)移项，得3(2x - 3) - 4(2x - 3) = 7 - 6.合并同类项，得-(2x - 3) = 1.去括号，得-2x + 3 = 1.移项、合并同类项，得-2x = -2.系数化为1，得x = 1.
(6)去分母，得8x - 160 = 5 - 4x.移项，得8x + 4x = 5 + 160.合并同类项，得12x = 165.系数化为1，得$x = \frac{55}{4}.$
(7)去分母，得4x - 1 = 6 - 2(3x - 2).去括号，得4x - 1 = 6 - 6x + 4.移项、合并同类项，得10x = 11.系数化为1，得$x = \frac{11}{10}. (8)$去括号，得2x - 8x + 8 - 8 = -3.移项，得2x - 8x = -3 - 8 + 8.合并同类项，得-6x = -3.系数化为1，得$x = \frac{1}{2}. (9)$去分母，得12 - 2(x - 1) = 4x - (11 + x).去括号，得12 - 2x + 2 = 4x - 11 - x.移项，得-2x - 4x + x = -11 - 2 - 12.合并同类项，得-5x = -25.系数化为1，得x = 5.
(10)方程可化为$\frac{10(x - 3)}{2} - \frac{10(x + 2)}{5} = 3，$即5(x - 3) - 2(x + 2) = 3.去括号，得5x - 15 - 2x - 4 = 3.移项，得5x - 2x = 3 + 15 + 4.合并同类项，得3x = 22.系数化为1，得$x = \frac{22}{3}.$
(11)去分母，得4(5 - x) = 6(x - 2) - 3(x + 1).去括号，得20 - 4x = 6x - 12 - 3x - 3.移项，得-4x - 6x + 3x = -12 - 3 - 20.合并同类项，得-7x = -35.系数化为1，得x = 5.
(12)去分母，得3(x + 1) - (x + 4) = 6 + 4x.去括号，得3x + 3 - x - 4 = 6 + 4x.移项，得3x - x - 4x = 6 + 4 - 3.合并同类项，得-2x = 7.系数化为1，得$x = -\frac{7}{2}.$