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МатематикаРешебник по алгебре 8 класс Макарычев ФГОС Задание 641
Задание 641
\[\boxed{\text{641\ (641).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ py - p - 1 = 0\]
\[py = p + 1\ \ \ \ \ |\ :p\]
\[y = \frac{p + 1}{p}\]
\[1.\ p = 0 \Longrightarrow y\ \varnothing\]
\[2.\ p \neq 0 \Longrightarrow y = \frac{p + 1}{p}\]
\[Ответ:при\ p \neq 0:\ \ \ y = \frac{p + 1}{p}.\]
\[\textbf{б)}\ py - 3y - 4p + 12 = 0\]
\[y(p - 3) = 4p - 12\ \ \ \ \ \ \ |\ :(p - 3)\]
\[y = \frac{4p - 12}{p - 3}\]
\[1.\ p - 3 = 0,\]
\[\ \ p = 3 \Longrightarrow y - любое\ число.\]
\[2.\ p - 3 \neq 0\ \Longrightarrow\]
\[\Longrightarrow y = \frac{4 \cdot (p - 3)}{p - 3} = 4\ \]
\[Ответ:при\ p = 0:\]
\[y - любое\ число;\]
\[при\ p \neq 3:\ \ y = 4.\]
\[\boxed{\text{641.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]
\[\textbf{а)}\ 1 + \frac{1}{3 + \frac{1}{2 + \frac{1}{5 - x^{2}}}} = 1\frac{7}{24}\]
\[1 + \frac{1}{3 + \frac{1}{\frac{10 - 2x^{2} + 1}{5 - x^{2}}}} = \frac{31}{24}\]
\[1 + \frac{1}{3 + \frac{5 - x^{2}}{11 - 2x^{2}}} = \frac{31}{24}\]
\[1 + \frac{1}{\frac{33 - 6x^{2} + 5 - x^{2}}{11 - 2x^{2}}} = \frac{31}{24}\]
\[1 \cdot \left( 38 - 7x^{2} \right) +\]
\[+ \frac{11 - 2x^{2}}{- 7x^{2} + 38} = \frac{31}{24}\]
\[\frac{- 7x^{2} + 38 + 11 - 2x^{2}}{- 7x^{2} + 38} = \frac{31}{24}\]
\[24 \cdot \left( - 9x^{2} + 49 \right) =\]
\[= 31 \cdot \left( - 7x^{2} + 38 \right)\]
\[- 216x^{2} + 1176 =\]
\[= - 217x^{2} + 1178\]
\[x^{2} = 2\]
\[x = \pm \sqrt{2}\]
\[Ответ:x = \pm \sqrt{2}.\]
\[\textbf{б)}\ 1 - \frac{1}{2 + \frac{1}{1 + \frac{1}{10 - x^{2}}}} = \frac{3}{5}\]
\[1 - \frac{1}{2 + \frac{1}{\frac{10 - x^{2} + 1}{10 - x^{2}}}} = \frac{3}{5}\]
\[1 - \frac{1}{2 + \frac{10 - x^{2}}{11 - x^{2}}} = \frac{3}{5}\]
\[1 - \frac{1}{\frac{22 - 2x^{2} + 10 - x^{2}}{11 - x^{2}}} = \frac{3}{5}\]
\[1 - \frac{11 - x^{2}}{32 - 3x^{2}} = \frac{3}{5}\]
\[\frac{32 - 3x^{2} - 11 + x^{2}}{32 - 3x^{2}} = \frac{3}{5}\]
\[5 \cdot \left( 21 - 2x^{2} \right) = 3 \cdot \left( 32 - 3x^{2} \right)\]
\[105 - 10x^{2} = 96 - 9x^{2}\]
\[x^{2} = 9\]
\[x = \pm 3\ \]
\[Ответ:x = \pm 3.\]
