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МатематикаРешебник по алгебре 8 класс Мерзляк ФГОС Задание 796
Задание 796
\[\boxed{\mathbf{796\ (796).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]
\[1)\ \frac{x^{2} - 8x + 7}{x - a} = 0\]
\[x^{2} - 8x + 7 = 0\]
\[x_{1} + x_{2} = 8,\ \ x_{1}x_{2} = 7,\]
\[\text{\ \ }x_{1} = 7,\ \ x_{2} = 1\]
\[x \neq a\]
\[\frac{(x - 7)(x - 1)}{x - a} = 0\]
\[если\ a = 1,\ \]
\[то\ \frac{(x - 7)(x - 1)}{x - 1} = 0,\ \ \]
\[x = 7,\ \ x \neq 1;\]
\[если\ a = 7,\ \]
\[то\ \frac{(x - 7)(x - 1)}{x - 7} = 0,\]
\[x = 1,\ \ x \neq 7;\ \]
\[если\ a \neq 7\ \ и\ \ \ a \neq 1,\ то\ x = 7\ \ \]
\[или\ \ x = 1.\]
\[2)\ \frac{x - a}{x^{2} - 8x + 7} = 0\]
\[\frac{x - a}{(x - 7)(x - 1)} = 0\]
\[если\ a \neq 7\ \ и\ \ \ a \neq 1,\ то\ x = a;\]
\[если\ a = 1\ \ или\ a = 7,\ \]
\[то\ корней\ нет.\]
\[3)\ \frac{x² - (3a + 2)x + 6a}{x - 6} = 0\]
\[x² - (3a + 2)x + 6a = 0\]
\[D = 9a^{2} + 12a + 4 - 24a =\]
\[= 9a^{2} - 12a + 4 = (3a - 2)²\]
\[x = \frac{3a + 2 + 3a - 2}{2} = 3a\]
\[x = \frac{3a + 2 - 3a + 2}{2} = 2\]
\[\frac{(x - 3a)(x - 2)}{x - 6} = 0\]
\[если\ a \neq 2\ \ и\ \ a \neq \frac{2}{3},\ то\ \ x = 3a\ \ \]
\[или\ x = 2,\ \ x \neq 6;\]
\[если\ a = 2\ \ или\ \ a = \frac{2}{3},то\ \ x = 2.\ \]
\[4)\ \frac{a(x - a)}{x + 3} = 0\]
\[если\ a = 0,\ то\ x - любое\ число,\ \]
\[кроме\ x = - 3;\]
\[если\ \ a = - 3,\ то\ корней\ нет,\]
\[так\ как\ a \neq 0;\]
\[если\ a \neq 0\ \ и\ \ a \neq - 3,\]
\[то\ \ x = a.\]
\[\boxed{\mathbf{7}\mathbf{9}\mathbf{6}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]
\[1)\ 2x^{2} - 5x + b;\ \ x - 3\]
\[x_{1} = 3\]
\[x_{1} + x_{2} = 2,5\]
\[x_{2} = 2,5 - x_{1} = 2,5 - 3 = - 0,5\]
\[x_{1} \cdot x_{2} = b\ :2\]
\[3 \cdot ( - 0,5) = - 1,5;\ \ \ \ \]
\[b = - 1,5 \cdot 2 = - 3\]
\[Ответ:\ b = - 3.\]
\[2) - 4x^{2} + bx + 2,\ \ x + 1\]
\[x_{1} = - 1\]
\[x_{1} \cdot x_{2} = - \frac{1}{2}\]
\[x_{2} = - \frac{1}{2}\ :x_{1} = - \frac{1}{2}\ :( - 1) = \frac{1}{2}\]
\[x_{1} + x_{2} = \frac{b}{4}\]
\[\frac{1}{2} - 1 = \frac{b}{4}\]
\[\frac{b}{4} = - \frac{1}{2}\]
\[2b = - 4\]
\[b = - 2\]
\[Ответ:\ b = - 2.\]
\[3)\ 3x² - 4x + b,\ \ \]
\[3x - 2 = 3 \cdot \left( x - \frac{2}{3} \right)\]
\[x_{1} = \frac{2}{3}\]
\[x_{1} + x_{2} = \frac{4}{3}\]
\[x_{2} = \frac{4}{3} - \frac{2}{3} = \frac{2}{3}\]
\[x_{1} \cdot x_{2} = \frac{b}{3}\]
\[\frac{2}{3} \cdot \frac{2}{3} = \frac{b}{3}\]
\[\frac{4}{9} = \frac{b}{3}\]
\[9b = 12\]
\[b = \frac{4}{3}\]
\[Ответ:b = \frac{4}{3}.\]
