помогите пожалуйста.. Created by aygh19635. matematika-ru

Ответ: 2.(решение на фото)

помогите пожалуйста....

Ответ:

$\frac{1}{8} \times ( \frac{4}{3} - x) + \frac{1}{3} = \frac{1}{4} \\ \frac{1}{8} \times \frac{4}{3} - \frac{1}{8} x + \frac{1}{3} = \frac{1}{4} \\ \frac{1}{6} - \frac{1}{8} x + \frac{1}{3} = \frac{1}{4} \\ - \frac{1}{8} x = \frac{1}{4} - \frac{1}{6} - \frac{1}{3} \\ \frac{1}{8} x = \frac{1}{6} + \frac{1}{3} - \frac{1}{4} \\ \frac{1}{8} x = \frac{2 + 4 - 3}{12} \\ \frac{1}{8} x = \frac{3}{12} \\ x = \frac{3}{12} \times \frac{8}{1} \\ x = 2$

Сделаем проверку:

$\frac{1}{8} ( \frac{4}{3} - 2) + \frac{1}{3} = \frac{1}{8} \times \frac{4}{3} - \frac{1}{8} \times 2 + \frac{1}{3} = \frac{1}{6} - \frac{1}{4} + \frac{1}{3} = \frac{2 - 3 + 4}{12} = \frac{3}{12} = \frac{1}{4}$

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