Đáp án + Giải thích các bước giải:

$\textbf{a$\bigg)$} \dfrac{x}{3} + 2x = 12$

$\dfrac{7x}{3} = 12$

$7x = 36$

$x = \dfrac{36}{7}$

Vậy $x = \dfrac{36}{7}$

$\textbf{b$\bigg)$} \dfrac{2x - 3}{5} = \dfrac{3x + 1}{3}$

$3(2x - 3) = 5(3x + 1)$

$6x - 9 = 15x + 5$

$-9x - 14 = 0$

$x = -\dfrac{14}{9}$

Vậy $x = -\dfrac{14}{9}$

$\textbf{c$\bigg)$} \dfrac{x - 2}{3} + \dfrac{2x + 1}{2} = 0$

$\dfrac{2(x - 2)}{6} + \dfrac{3(2x + 1)}{6} = 0$

$2x - 4 + 6x + 3 = 0$

$8x - 1 = 0$

$x = \dfrac{1}{8}$

Vậy $x = \dfrac{1}{8}$

$\textbf{d$\bigg)$} \dfrac{x - 3}{7} + \dfrac{x + 1}{2} - \dfrac{x}{4}= 0$

$\dfrac{4(x - 3)}{28} + \dfrac{14(x + 1)}{28} - \dfrac{7x}{28} = 0$

$4x - 12 + 14x + 14 - 7x = 0$

$11x + 2 = 0$

$x = -\dfrac{2}{11}$

Vậy $x= -\dfrac{2}{11}$

$\textbf{e$\bigg)$} \dfrac{3x + 2}{2} + \dfrac{5 - 2x}{3} = \dfrac{11}{12}$

$\dfrac{6(3x + 2)}{12} + \dfrac{4(5 - 2x)}{12} = \dfrac{11}{12}$

$18x + 12 + 20 - 8x = 11$

$10x + 21 = 0$

$x = -\dfrac{21}{10}$

Vậy $x = -\dfrac{21}{10}$

$\textbf{f$\bigg)$} \dfrac{x - 2}{4} + \dfrac{x - 3}{8} = \dfrac{2x - 3}{6}$

$\dfrac{6(x - 2)}{24} + \dfrac{3(x - 3)}{24} = \dfrac{4(2x - 3)}{24}$

$6x - 12 + 3x - 9 = 8x - 12$

$x - 9 = 0$

$x = 9$

Vậy $x = 9$

$\textbf{g$\bigg)$} \dfrac{2x - 1}{5} - \dfrac{x - 1}{4} + \dfrac{x + 1}{10} = 1$

$\dfrac{4(2x - 1)}{20} - \dfrac{5(x -1)}{20} + \dfrac{2(x + 1)}{20} = \dfrac{20}{20}$

$8x - 4 - 5x + 5 + 2x + 2 = 20$

$5x - 17 = 0$

$x = \dfrac{17}{5}$

Vậy $x = \dfrac{17}{5}$

$\textbf{h$\bigg)$} 3x - \dfrac{2x - 1}{5} = 3 + \dfrac{3 - x}{4}$

$\dfrac{15x}{5} - \dfrac{2x - 1}{5} = \dfrac{12 + 3 - x}{4}$

$\dfrac{13x + 1}{5} = \dfrac{15 - x}{4}$

$4(13x + 1) = 5(15 - x)$

$52x + 4 = 75 - 5x$

$57x = 71$

$x = \dfrac{71}{57}$

Vậy $x = \dfrac{71}{57}$

$\textbf{i$\bigg)$} \dfrac{x - 5}{15} + \dfrac{x - 5}{14} =\dfrac{x - 5}{13} + \dfrac{x -5}{12}$

$\dfrac{x - 5}{15} + \dfrac{x - 5}{14} - \dfrac{x - 5}{13} - \dfrac{x -5}{12} = 0$

$(x - 5)\bigg(\dfrac{1}{15} + \dfrac{1}{14} - \dfrac{1}{13} - \dfrac{1}{12}\bigg) = 0$

Vì $\dfrac{1}{15} + \dfrac{1}{14} - \dfrac{1}{13} - \dfrac{1}{12} \ne 0$ nên $x - 5 = 0$, hay $x = 5$.

Vậy $x = 5$

$\textbf{k$\bigg)$} \dfrac{x - 105}{5} + \dfrac{x - 103}{7} + \dfrac{x - 101}{9} + \dfrac{x - 99}{11} = 4$

$\dfrac{x - 105}{5} - 1 + \dfrac{x - 103}{7} - 1 + \dfrac{x - 101}{9} - 1 + \dfrac{x - 99}{11} - 1 = 0$

$\dfrac{x - 110}{5} + \dfrac{x - 110}{7} + \dfrac{x - 110}{9} + \dfrac{x - 110}{11} = 0$

$(x - 110)\bigg(\dfrac{1}{5} + \dfrac{1}{7} - \dfrac{1}{9} - \dfrac{1}{11}\bigg) = 0$

Vì $\dfrac{1}{5} + \dfrac{1}{7} - \dfrac{1}{9} - \dfrac{1}{11} \ne 0$ nên $x - 110 =0$, hay $x = 110$.

Vậy $x = 110$