Решебник по алгебре 8 класс Мерзляк Задание 759

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\[1)\ 2x^{2} - 5x + b;\ \ x - 3\]

\[x_{1} = 3\]

\[x_{1} + x_{2} = 2,5\]

\[x_{2} = 2,5 - x_{1} = 2,5 - 3 = - 0,5\]

\[x_{1} \cdot x_{2} = b\ :2\]

\[3 \cdot ( - 0,5) = - 1,5;\ \ \ \ \]

\[b = - 1,5 \cdot 2 = - 3\]

\[Ответ:\ b = - 3.\]

\[2) - 4x^{2} + bx + 2,\ \ x + 1\]

\[x_{1} = - 1\]

\[x_{1} \cdot x_{2} = - \frac{1}{2}\]

\[x_{2} = - \frac{1}{2}\ :x_{1} = - \frac{1}{2}\ :( - 1) = \frac{1}{2}\]

\[x_{1} + x_{2} = \frac{b}{4}\]

\[\frac{1}{2} - 1 = \frac{b}{4}\]

\[\frac{b}{4} = - \frac{1}{2}\]

\[2b = - 4\]

\[b = - 2\]

\[Ответ:\ b = - 2.\]

\[3)\ 3x² - 4x + b,\ \ \]

\[3x - 2 = 3 \cdot \left( x - \frac{2}{3} \right)\]

\[x_{1} = \frac{2}{3}\]

\[x_{1} + x_{2} = \frac{4}{3}\]

\[x_{2} = \frac{4}{3} - \frac{2}{3} = \frac{2}{3}\]

\[x_{1} \cdot x_{2} = \frac{b}{3}\]

\[\frac{2}{3} \cdot \frac{2}{3} = \frac{b}{3}\]

\[\frac{4}{9} = \frac{b}{3}\]

\[9b = 12\]

\[b = \frac{4}{3}\]

\[Ответ:b = \frac{4}{3}.\]