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МатематикаРешебник по алгебре 8 класс Макарычев ФГОС Задание 632
Задание 632
\[\boxed{\text{632\ (632).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[Пусть\ \text{x\ }ч - потребуется\ \]
\[второму\ крану,\ тогда\ \]
\[(x + 5)\ ч - первому\ крану.\ \]
\[Известно,\ что\ при\ совместной\ \]
\[разгрузке\ работа\ закончилась\]
\[через\ 6\ часов.\ \]
\[Составим\ уравнение:\]
\[1\ :\left( \frac{1}{x} + \frac{1}{x + 5} \right) = 6\]
\[1\ :\left( \frac{x + 5 + x}{x^{2} + 5x} \right) = 6\]
\[\frac{x^{2} + 5x}{2x + 5} = 6\]
\[x^{2} + 5x = 6 \cdot (2x + 5)\]
\[x^{2} + 5x - 12x - 30 = 0\]
\[x^{2} - 7x - 30 = 0\]
\[D = 49 + 120 = 169\]
\[x_{2} = \frac{7 + 13}{2} = 10\ (ч) -\]
\[потребуется\ второму\ крану.\]
\[x + 5 = 10 + 5 = 15\ (ч) -\]
\[потребуется\ первому\ крану.\]
\[Ответ:15\ ч\ и\ 10\ ч.\ \]
\[\boxed{\text{632.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]
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\[\textbf{а)}\ \frac{2x - 5}{x + 5} - 4 = 0\ \ \ \ \ \ | \cdot (x + 5)\] \[x \neq - 5\] \[2x - 5 - 4 \cdot (x + 5) = 0\] \[2x - 5 - 4x - 20 = 0\] \[- 2x - 25 = 0\] \[- 2x = 25\] \[x = - \frac{25}{2}\] \[x = - 12,5\] \[Ответ:\ x = - 12,5.\] |
\[\textbf{д)}\ \frac{8}{x} = 3x + 2\ \ \ \ \ \ | \cdot x,\ \ x \neq 0\] \[8 = x \cdot (3x + 2)\] \[8 = 3x^{2} + 2x\] \[3x^{2} + 2x - 8 = 0\] \[D = 4 + 4 \cdot 3 \cdot 8 = 4 + 96 = 100\] \[x_{1,2} = \frac{- 2 \pm \sqrt{100}}{2 \cdot 3} = \frac{- 2 \pm 10}{6}\] \[x_{1} = - 2;\ \ x_{2} = \frac{8}{6} = \frac{4}{3} = 1\frac{1}{3}\] \[Ответ:\ x = - 2,\ x = 1\frac{1}{3}.\] |
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\[\textbf{б)}\frac{12}{7 - x} = x\ \ \ \ \ \ \ | \cdot (7 - x)\] \[x \neq 7\] \[12 = x(7 - x)\] \[12 = 7x - x^{2}\] \[x^{2} - 7x + 12 = 0\] \[D = 49 - 4 \cdot 12 = 49 - 48 = 1\] \[x_{1,2} = \frac{7 \pm \sqrt{1}}{2} = \frac{7 \pm 1}{2}\] \[x_{1} = 4;\ \ x_{2} = 3\] \[Ответ:\ x = 3,\ x = 4\text{.\ }\] |
\[\textbf{е)}\frac{x^{2} + 4x}{x + 2} = \frac{2x}{3}\ \ \ \ \ \ | \cdot 3 \cdot (x + 2)\] \[x \neq - 2\] \[3 \cdot \left( x^{2} + 4x \right) = 2x \cdot (x + 2)\] \[3x^{2} + 12x = 2x^{2} + 4x\] \[x^{2} + 8x = 0\] \[x \cdot (x + 8) = 0\] \[x = 0,\ \ x = - 8\] \[Ответ:\ x = 0,\ x = - 8.\] |
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\[\textbf{в)}\ \frac{x^{2} - 4}{4x} = \frac{3x - 2}{2x}\ \ \ \ \ \ \ \ \ \ | \cdot 4x\] \[x \neq 0\] \[x^{2} - 4 = 2 \cdot (3x - 2)\] \[x^{2} - 4 = 6x - 4\] \[x^{2} - 6x = 0\] \[x(x - 6) = 0\] \[x \neq 0,\ \ x = 6\] \[Ответ:\ x = 6.\] |
\[\textbf{ж)}\ \frac{2x^{2} - 5x + 3}{10x - 5} = 0\ \ \ | \cdot (10x - 5)\] \[x \neq 0,5\] \[2x^{2} - 5x + 3 = 0\] \[D = 25 - 4 \cdot 2 \cdot 3 = 25 - 24 = 1\] \[x_{1,2} = \frac{5 \pm \sqrt{1}}{2 \cdot 2} = \frac{5 \pm 1}{4}\] \[x_{1} = \frac{6}{4} = \frac{3}{2} = 1,5\] \[x_{2} = 1\] \[Ответ:\ x = 1;x = 1,5.\] |
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\[\textbf{г)}\frac{10}{2x - 3} = x - 1\ \ \ \ \ \ | \cdot (2x - 3)\] \[x \neq 1,5\] \[10 = (2x - 3)(x - 1)\] \[10 = 2x^{2} - 2x - 3x + 3\] \[2x^{2} - 5x + 3 - 10 = 0\] \[2x^{2} - 5x - 7 = 0\] \[D = 25 - 4 \cdot 2 \cdot ( - 7) = 25 + 56 = 81\] \[x_{1,2} = \frac{5 \pm \sqrt{81}}{2 \cdot 2} = \frac{5 \pm 9}{4}\] \[x_{1} = - 1;\ \ x_{2} = \frac{14}{4} = \frac{7}{2} = 3,5\] \[Ответ:x = - 1,\ x = 3,5.\ \] |
\[\textbf{з)}\ \frac{4x^{3} - 9x}{x + 1,5} = 0\ \ \ \ \ \ \ \ | \cdot (x + 1,5)\] \[x \neq - 1,5\] \[4x^{3} - 9x = 0\] \[x \cdot \left( 4x^{2} - 9 \right) = 0\] \[x = 0,\ \ 4x^{2} = 9\] \[\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }x^{2} = \frac{9}{4}\] \[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 1,5\] \[Ответ:x = 0,\ x = 1,5.\ \] |
