Question

Решите уравнение пожалуйста

Answer 1

$\tt\displaystyle 1) \; \dfrac{\dfrac{1}{6}}{\dfrac{1}{4} x - \dfrac{1}{3}} - \dfrac{3}{8} = \dfrac{1}{40} \\\\ \dfrac{1}{6} : (\dfrac{1}{4} x - \dfrac{1}{3}) = \dfrac{1}{40} + \dfrac{3}{8} \\\\ \dfrac{1}{6} : (\dfrac{3}{12} x - \dfrac{4}{12}) = \dfrac{1+15}{40} \\\\ \dfrac{1}{6} : \dfrac{3x-4}{12} = \dfrac{16}{40} \\\\ \dfrac{1}{6}\cdot \dfrac{12}{3x-4} = \dfrac{2}{5} \\\\ \dfrac{2}{3x-4} = 0,4 \\\\ 2 : (3x - 4) = 0,4 \\ 3x - 4 = 2 : 0,4 \\ 3x - 4 = 5 \\ 3x = 5 + 4 \\ 3x = 9 \\ x = 9 : 3 \\ x = 3 \\\\\\ 2) \; \dfrac{\dfrac{1}{4}}{\dfrac{3}{4} x - \dfrac{1}{2}} + \dfrac{2}{5} = \dfrac{1}{2} \\\\ (\dfrac{1}{4} : (\dfrac{3}{4} x - \dfrac{1}{2})) = \dfrac{1}{2} - \dfrac{2}{5} \\\\ (\dfrac{1}{4} : (\dfrac{3}{4} x = \dfrac{2}{4})) = \dfrac{5-4}{10} \\\\ \dfrac{1}{4} : \dfrac{3x-2}{4} = \dfrac{1}{10} \\\\ \dfrac{1}{4}\cdot \dfrac{4}{3x-2} = \dfrac{1}{10} \\\\ \dfrac{1}{3x-2} = \dfrac{1}{10} \\\\3x-2 = 10 \\3x = 10+2 \\ 3x = 12 \\ x = 12 : 3 \\ x = 4$