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МатематикаГДЗ по алгебре 8 класс Макарычев ФГОС Задание 633
Задание 633
\[\boxed{\text{633\ (633).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[Пусть\ x\ ч - время\ работы\ \]
\[второго\ автомата,\ тогда\ \]
\[(x + 2)\ ч - время\ первого\ \]
\[автомата.\ Оба\ автомата\ могут\ \]
\[изготовить\ это\ количество\]
\[деталей\ за\ 2\ ч\ 55\ мин =\]
\[= 2\frac{11}{12}\ ч = \frac{35}{12}\ ч.\]
\[Составим\ уравнение:\]
\[\frac{1}{x + 2} + \frac{1}{x} = \frac{12}{35}\ \ \ \ \ \ \ \ | \cdot 35x(x + 2)\]
\[35x + 35x + 70 = 12x^{2} + 24x\]
\[12x^{2} - 46x - 70 = 0\ \ \ \ \ \ |\ :2\]
\[6x^{2} - 23x - 35 = 0\]
\[D = 529 + 840 = 1369 = 37^{2}\]
\[x_{1,2} = \frac{23 \pm 37}{12} = \frac{60}{12};\ - \frac{14}{12} <\]
\[< 0 - не\ подходит\]
\[x = 5\ (ч) - время\ работы\ \]
\[второго\ автомата.\]
\[x + 2 = 5 + 2 = 7\ (ч) - время\]
\[\ работы\ первого\ автомата.\]
\[Ответ:7\ ч.\ \]
\[\boxed{\text{633.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]
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\[\textbf{а)}\frac{x^{2}}{x^{2} + 1} = \frac{7x}{x^{2} + 1}\ \ \ \ \ \ \ | \cdot \left( x^{2} + 1 \right)\] \[x^{2} = 7x\] \[x^{2} - 7x = 0\] \[x(x - 7) = 0\] \[x = 0,\ \ x = 7\] \[Ответ:x = \left\{ 0;7 \right\}.\] |
\[\textbf{б)}\frac{y^{2}}{y^{2} - 6y} = \frac{4 \cdot (3 - 2y)}{y \cdot (6 - y)}\] \[при\ y^{2} - 6y \neq 0\ \ \ \] \[y(y - 6) \neq 0\] \[y \neq 0,\ \ y \neq 6\] \[\frac{y^{2}}{y^{2} - 6y} = \frac{4 \cdot (3 - 2y)}{6y - y^{2}}\] \[\frac{y^{2}}{y^{2} - 6y} = \frac{- 4 \cdot (3 - 2y)}{y^{2} - 6y}\ \ \ \ \ \ | \cdot \left( y^{2} - 6y \right)\] \[y^{2} = - 12 + 8y\] \[y^{2} - 8y + 12 = 0\] \[D = 64 - 48 = 16\] \[x_{1,2} = \frac{8 \pm 4}{2} = 6;2,\ \ (y \neq 6)\] \[Ответ:y = 2.\] |
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\[\textbf{в)}\frac{x - 2}{x + 2} = \frac{x + 3}{x - 4}\] \[при\ x + 2 \neq 0\ \ \ и\ \ \ x - 4 \neq 0\] \[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \neq - 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ x \neq 4\] \[(x - 2)(x - 4) = (x + 3)(x + 2)\] \[x^{2} - 4x - 2x + 8 = x^{2} + 2x + 3x + 6\] \[- 11x = - 2\] \[x = \frac{2}{11}\] \[Ответ:x = \frac{2}{11}.\] |
\[\textbf{г)}\frac{8y - 5}{y} = \frac{9y}{y + 2}\] \[при\ y \neq 0\ \ \ и\ \ \ y + 2 \neq 0\] \[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y \neq - 2\] \[(8y - 5)(y + 2) = 9y \cdot y\] \[8y^{2} + 16y - 5y - 10 - 9y^{2} = 0\] \[y^{2} - 11y + 10 = 0\] \[D = 121 - 40 = 81\] \[y_{1,2} = \frac{11 \pm 9}{2} = 10;1\] \[Ответ:y = \left\{ 1;10 \right\}.\] |
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\[\textbf{д)}\frac{x^{2} + 3}{x^{2} + 1} = 2\ \ \ \ \ | \cdot \left( x^{2} + 1 \right)\] \[x^{2} + 3 = 2 \cdot \left( x^{2} + 1 \right)\] \[x^{2} + 3 = 2x^{2} + 2\] \[x^{2} - 1 = 0\] \[x^{2} = 1\] \[x = \pm 1\] \[Ответ:x = \left\{ - 1;\ 1 \right\}.\] |
\[\textbf{е)}\frac{3}{x^{2} + 2} = \frac{1}{x},\ \ \] \[при\ x \neq 0\] \[3x = x^{2} + 2\] \[x^{2} - 3x + 2 = 0\] \[D = 9 - 8 = 1\] \[x_{1,2} = \frac{3 \pm 1}{2} = 2;1.\] \[Ответ:x = \left\{ 1;2 \right\}\text{.\ }\] |
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\[\textbf{ж)}\ x + 2 = \frac{15}{4x + 1}\] \[при\ 4x + 1 \neq 0\] \[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \neq - \frac{1}{4}\] \[(x + 2)(4x + 1) = 15\] \[4x^{2} + x + 8x + 2 = 15\] \[4x^{2} + 9x - 13 = 0\] \[D = 81 + 208 = 289 = 17^{2}\] \[x_{1,2} = \frac{- 9 \pm 17}{8} = 1;\ - \frac{13}{4}\] \[Ответ:x = \left\{ - 3,25;1 \right\}.\] |
\[\textbf{з)}\frac{x^{2} - 5}{x - 1} = \frac{7x + 10}{9}\] \[при\ x - 1 \neq 0\] \[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \neq 1\] \[9 \cdot \left( x^{2} - 5 \right) = (7x + 10)(x - 1)\] \[9x^{2} - 45 = 7x^{2} + 3x - 10\] \[2x^{2} - 3x - 35 = 0\] \[D = 9 + 280 = 289 = 17^{2}\] \[x_{1,2} = \frac{3 \pm 17}{4} = 5;\ - 3,5\] \[Ответ:x = \left\{ - 3,5;5 \right\}\text{.\ \ \ }\] |
