Решебник по алгебре 8 класс Мерзляк Задание 785

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\[1)\frac{2y}{y - 3} = \frac{3y + 3}{y}\]

\[\frac{2y}{y - 3} - \frac{3y + 3}{y} = 0\]

\[\frac{2y^{2} - 3y^{2} + 6y + 9}{y(y - 3)} = 0;\ \ \ \ \]

\[y \neq 0;\ \ \ \ y \neq 3\]

\[D = 36 + 36 = 72\]

\[y_{1,2} = \frac{- 6 \pm 6\sqrt{2}}{- 2} = 3 \pm 3\sqrt{2}\]

\[Ответ:\ y = 3 \pm 3\sqrt{2}.\]

\[2)\ \frac{3x + 4}{x - 2} = \frac{2x - 9}{x + 1}\]

\[\frac{3x + 4}{x - 2} - \frac{2x - 9}{x + 1} = 0\]

\[x^{2} + 22x - 33 = 0.\]

\[x_{1} + x_{2} = - 22;\ \ \ \ \ x_{1}x_{2} = - 23,\ \ \]

\[x_{1} = - 23,\ \ x_{2} = 1\]

\[Ответ:\ x = - 23;x = 1.\]

\[3)\ \frac{5x + 2}{x - 1} = \frac{4x + 13}{x + 7}\]

\[\frac{5x + 2}{x - 1} - \frac{4x + 13}{x + 7} = 0\]

\[\ x_{1} + x_{2} = - 28;\ \ \ \ \ x_{1}x_{2} = 27,\ \ \]

\[x_{1} = - 27,\ \ x_{2} = \ - 1\ \]

\[Ответ:\ x = - 27;\ x = - 1.\]

\[4)\ \frac{2x^{2} - 3x + 1}{x - 1} = 3x - 4;\ \ \ \ \ \ \ \ \ \]

\[x \neq 1\]

\[- x^{2} + 4x - 3 = 0\]

\[x^{2} - 4x + 3 = 0\]

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

\[x_{1} = 3,\ \ \]

\[x_{2} = 1\ (не\ подходит)\]

\[Ответ:x = 3.\]