ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 327

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Пояснение.

Решение.

\[\textbf{а)}\ \frac{x - |x - 1|}{x + 2} = ?\]

\[x = 4:\]

\[\frac{4 - |4 - 1|}{4 + 2} = \frac{4 - 3}{6} = \frac{1}{6}.\]

\[x = 38:\]

\[\frac{38 - |38 - 1|}{38 + 2} = \frac{38 - 37}{40} = \frac{1}{40}.\]

\[x = - 42:\]

\[\frac{- 42 - | - 42 - 1|}{- 42 + 2} = \frac{- 42 - 43}{- 40} =\]

\[= - \frac{85}{- 40} = \frac{85}{40} = \frac{17}{8} = 2\frac{1}{8}.\]

\[\textbf{б)}\ \frac{2|3 - x| - 1}{4} = ?\]

\[x = 2:\]

\[\frac{2|3 - 2| - 1}{4} = \frac{2 - 1}{4} = \frac{1}{4}.\]

\[x = 11:\]

\[\frac{2|3 - 11| - 1}{4} = \frac{2 \cdot ( - 8) - 1}{4} =\]

\[= - \frac{17}{4} = - 4\frac{1}{4}.\]

\[x = - 6:\]

\[\frac{2|3 + 6| - 1}{4} = \frac{18 - 1}{4} = \frac{17}{4} =\]

\[= 4\frac{1}{4}.\]