Giải thích các bước giải:

$\begin{align*}
& \text{Câu 1: } \\
&a) \dfrac{-5}{12} + \dfrac{4}{39} + \dfrac{17}{12} - \dfrac{43}{39} = \left(\dfrac{-5}{12} + \dfrac{17}{12}\right) + \left(\dfrac{4}{39} - \dfrac{43}{39}\right) \\
& = \dfrac{12}{12} + \dfrac{-39}{39} = 1 - 1 = 0 \\
&b) -0,25 + \dfrac{2}{5} : \dfrac{16}{5} = -\dfrac{1}{4} + \dfrac{2}{5} \cdot \dfrac{5}{16} \\
& = -\dfrac{1}{4} + \dfrac{1}{8} = -\dfrac{2}{8} + \dfrac{1}{8} = -\dfrac{1}{8} \\
& \text{c) Ta có:} \\
& 12^2 + 5 \cdot \sqrt{49} = 144 + 5 \cdot 7 \\
& = 144 + 35 = 179 \\
& d) \dfrac{5}{7} \cdot \left(\dfrac{-2}{9}\right) + \dfrac{9}{7} \cdot \left(\dfrac{-2}{9}\right) = \left(\dfrac{-2}{9}\right) \cdot \left(\dfrac{5}{7} + \dfrac{9}{7}\right) \\
& = \dfrac{-2}{9} \cdot \dfrac{14}{7} = \dfrac{-2}{9} \cdot 2 = -\dfrac{4}{9} \\
& \text{h) Ta có:} \\
& \left(\dfrac{-1}{3}\right)^2 - \dfrac{3}{8} : (0,5)^3 - \dfrac{5}{2} \cdot (-4) + 5075^0 = \dfrac{1}{9} - \dfrac{3}{8} : \dfrac{1}{8} - (-10) + 1 \\
& = \dfrac{1}{9} - \dfrac{3}{8} \cdot 8 + 10 + 1 = \dfrac{1}{9} - 3 + 11 = \dfrac{1}{9} + 8 = \dfrac{73}{9} \\
\\
& \text{Câu 2:} \\
& \text{a) Ta có:} \\
& \dfrac{-3}{7} x = \dfrac{1}{2} - \dfrac{1}{3} \Leftrightarrow \dfrac{-3}{7} x = \dfrac{1}{6} \\
& x = \dfrac{1}{6} : \left(\dfrac{-3}{7}\right) = \dfrac{1}{6} \cdot \left(\dfrac{-7}{3}\right) = -\dfrac{7}{18} \\
& \text{b) Ta có:} \\
& \left(\dfrac{2}{5} - 3x\right)^2 = \dfrac{9}{25} \Leftrightarrow \dfrac{2}{5} - 3x = \dfrac{3}{5} \text{ hoặc } \dfrac{2}{5} - 3x = -\dfrac{3}{5} \\
& \text{Trường hợp 1:} \\
& \dfrac{2}{5} - 3x = \dfrac{3}{5} \Leftrightarrow 3x = \dfrac{2}{5} - \dfrac{3}{5} = -\dfrac{1}{5} \Leftrightarrow x = -\dfrac{1}{15} \\
& \text{Trường hợp 2:} \\
& \dfrac{2}{5} - 3x = -\dfrac{3}{5} \Leftrightarrow 3x = \dfrac{2}{5} - \left(-\dfrac{3}{5}\right) = 1 \Leftrightarrow x = \dfrac{1}{3} \\
& \text{c) Ta có:} \\
& \left(x - \dfrac{3}{5}\right) : \dfrac{-1}{3} = 0,4 \Leftrightarrow x - \dfrac{3}{5} = \dfrac{2}{5} \cdot \left(\dfrac{-1}{3}\right) \\
& \text{Thu gọn và tìm x:} \\
& x - \dfrac{3}{5} = -\dfrac{2}{15} \Leftrightarrow x = -\dfrac{2}{15} + \dfrac{9}{15} = \dfrac{7}{15} \\
& \text{d) Ta có:} \\
& (3x+1)^2 = 0 \Leftrightarrow 3x + 1 = 0 \\
& 3x = -1 \Leftrightarrow x = -\dfrac{1}{3}
\end{align*}$