ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 796

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\[\textbf{а)}\ \frac{2m^{2} - 8}{m^{2} + 6m + 8}\]

\[m² + 6m + 8 = 0\]

\[D_{1} = 3^{2} - 8 = 9 - 8 = 1\]

\[m_{1} = - 3 - 1 = - 4;\ \ \]

\[m_{2} = - 3 + 1 = - 2;\]

\[m^{2} + 6m + 8 = (m + 4)(m + 2);\]

\[\frac{2m^{2} - 8}{m^{2} + 6m + 8} =\]

\[= \frac{2 \cdot \left( m^{2} - 4 \right)}{(m + 4)(m + 2)} =\]

\[= \frac{2 \cdot (m - 2)(m + 2)}{(m + 4)(m + 2)} =\]

\[= \frac{2 \cdot (m - 2)}{(m + 4)}.\]

\[\textbf{б)}\ \frac{2m^{2} - 5m + 2}{mn - 2n - 3m + 6}\]

\[2m² - 5m + 2 = 0\]

\[D = 25 - 4 \cdot 2 \cdot 2 =\]

\[= 25 - 16 = 9\]

\[m_{1} = \frac{5 + 3}{4} = 2;\ \ \]

\[m_{2} = \frac{5 - 3}{4} = \frac{1}{2};\]

\[2m^{2} - 5m + 2 =\]

\[= 2 \cdot (m - 2)\left( m - \frac{1}{2} \right) =\]

\[= (m - 2)(2m - 1);\]

\[\frac{2m^{2} - 5m + 2}{mn - 2n - 3m + 6} =\]

\[= \frac{(m - 2)(2m - 1)}{n(m - 2) - 3 \cdot (m - 2)} =\]

\[= \frac{(m - 2)(2m - 1)}{(m - 2)(n - 3)} =\]

\[= \frac{2m - 1}{n - 3}.\]