ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 58

 

\[\boxed{\text{58\ (58).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \frac{(a + b)^{2}}{\text{ab}} - \frac{(a - b)^{2}}{\text{ab}} = 4\]

\[\frac{a^{2} + 2ab + b^{2}}{\text{ab}} - \frac{a^{2} - 2ab + b^{2}}{\text{ab}} =\]

\[= 4\]

\[\frac{a^{2} + 2ab + b^{2} - a^{2} + 2ab - b^{2}}{\text{ab}} =\]

\[= 4\]

\[\frac{4\text{ab}}{\text{ab}} = 4\]

\[4 = 4\]

\[Что\ и\ требовалось\ доказать.\ \]

\[\textbf{б)}\ \frac{(a + b)^{2}}{a^{2} + b^{2}} + \frac{(a - b)^{2}}{a^{2} + b^{2}} = 2\]

\[\frac{a^{2} + 2ab + b^{2}}{a^{2} + b^{2}} + \frac{a^{2} - 2ab + b^{2}}{a^{2} + b^{2}} =\]

\[= 2\]

\[\frac{a^{2} + 2ab + b^{2} + a^{2} - 2ab + b^{2}}{a^{2} + b^{2}} =\]

\[= 2\]

\[\frac{{2a}^{2} + {2b}^{2}}{a^{2} + b^{2}} = 2\]

\[2 = 2\]

\[Что\ и\ требовалось\ доказать.\]

\[\boxed{\text{58.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \frac{17 - 12x}{x} + \frac{10 - x}{x} =\]

\[= \frac{17 - 12x + 10 - x}{x} =\]

\[= \frac{27 - 13x}{x}\]

\[\textbf{б)}\ \frac{12p - 1}{3p^{2}} - \frac{1 - 3p}{3p^{2}} =\]

\[= \frac{12p - 1 - 1 + 3p}{3p^{2}} = \frac{15p - 2}{3p^{2}}\]

\[\textbf{в)}\ \frac{6y - 3}{5y} - \frac{y + 2}{5y} =\]

\[= \frac{6y - 3 - y - 2}{5y} = \frac{5y - 5}{5y} =\]

\[\textbf{г)}\ \frac{3p - q}{5p} - \frac{2p + 6q}{5p} + \frac{p - 4q}{5p} =\]

\[= \frac{3p - q - 2p - 6q + p - 4q}{5p} =\]

\[= \frac{2p - 11q}{5p}\]

\[\textbf{д)}\ \frac{5c - 2d}{4c} - \frac{3d}{4c} + \frac{d - 5c}{4c} =\]

\[= \frac{5c - 2d - 3d + d - 5c}{4c} =\]

\[\textbf{е)}\ \frac{2a}{b} - \frac{1 - 6a}{b} + \frac{13 - 8a}{b} =\]

\[= \frac{2a - 1 + 6a + 13 - 8a}{b} = \frac{12}{b}\]