Решебник по алгебре 9 класс Макарычев Задание 227

 

\[\boxed{\text{227\ (227).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

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

\[m^{2} + 6m + 8 = 0\]

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

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

\[= - 3 + 1 = - 2;\]

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

\[= (m + 4)(m + 2);\]

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

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

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

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

\[2m^{2} - 5m + 2 = 0\]

\[D = 25 - 4 \cdot 2 \cdot 2 = 25 - 16 = 9\]

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

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

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

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

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

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

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

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

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

\[\boxed{\text{227.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[x^{4} - 1 - 4x^{2} + 44 = 0\]

\[x^{4} - 4x^{2} + 43 = 0\]

\[Пусть\ x^{2} = t,\ \ x^{4} = t^{2},\]

\[\ \ t \geq 0,\ тогда:\]

\[t^{2} - 4t + 43 = 0\]

\[D = 4 - 43 < 0 \Longrightarrow корней\ нет.\]

\[Ответ:нет\ корней.\]

\[3x^{2}\left( x^{2} - 1 \right) - 10x^{2} + 4 = 0\]

\[3x^{4} - 3x^{2} - 10x^{2} + 4 = 0\]

\[3x^{4} - 13x^{2} + 4 = 0\]

\[Пусть\ x^{2} = t,\ \ x^{4} = t^{2},\]

\[\ \ t \geq 0,\ тогда:\]

\[3t^{2} - 13t + 4 = 0\]

\[D = 13^{2} - 4 \cdot 3 \cdot 4 =\]

\[= 169 - 48 = 121\]

\[t_{1,2} = \frac{13 \pm 11}{6},\ \ t_{1} = 4,\ \ \]

\[t_{2} = \frac{1}{3}.\]

\[\left\{ \begin{matrix} x² = 4 \\ x² = \frac{1}{3}\ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x = \pm 2 \\ x = \pm \sqrt{\frac{1}{3}} \\ \end{matrix} \right.\ .\]

\[Ответ:x = \pm 2;\ \ x = \pm \sqrt{\frac{1}{3}}.\]