Решебник по алгебре 8 класс Макарычев ФГОС Задание 333

 

\[\boxed{\text{333\ (}\text{с}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\frac{|x|}{x} = ?\]

\[x = - 8:\ \ \ \ \ \ \ \ \ \ \]

\[\frac{| - 8|}{- 8} = \frac{8}{- 8} = - 1.\]

\[x = - 5:\ \ \ \ \ \ \]

\[\frac{| - 5|}{- 5} = \frac{5}{- 5} = - 1.\]

\[x = 1:\ \ \ \ \ \ \ \ \ \]

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

\[x = 7:\ \text{\ \ \ \ \ \ \ \ }\]

\[\frac{|7|}{7} = \frac{7}{7} = 1.\]

\[x = 128:\ \ \ \ \]

\[\frac{|128|}{128} = \frac{128}{128} = 1.\]

\[\textbf{а)}\ \ при\ x > 0:\ \ \ \ \ \ \ \]

\[\frac{|x|}{x} = \frac{x}{x} = 1.\]

\[\textbf{б)}\ при\ x < 0:\ \ \]

\[\frac{|x|}{x} = - \frac{x}{x} = - 1.\ \]

\[\boxed{\text{333\ (}\text{н}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

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

\[x = 4:\]

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

\[x = 38:\]

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

\[x = - 42:\]

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

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

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

\[x = 2:\]

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

\[x = 11:\]

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

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

\[x = - 6:\]

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

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

\[\boxed{\text{333.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[ед.\ отрезок = 1\ клетка.\]

\[Отмечено\ число\ больше\ 5,\ но\ \]

\[меньше\ 6.\]

\[\frac{142}{29} = 4\frac{26}{29};\]

\[\sqrt{33} \approx 5,7.\]

\[A\left( \sqrt{33} \right).\]