Решебник по алгебре 9 класс Макарычев Задание 228

 

\[\boxed{\text{228\ (228).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \frac{x + 4}{x - 1} - \frac{37x - 12}{4x^{2} - 3x - 1}\text{\ \ }\]

\[4x^{2} - 3x - 1 = 0\]

\[D = 9 + 4 \cdot 4 = 9 + 16 = 25\]

\[x_{1} = \frac{3 + 5}{8} = 1;\ \ \ x_{2} = \frac{3 - 5}{8} =\]

\[= - \frac{1}{4} \Longrightarrow\]

\[4x^{2} - 3x - 1 =\]

\[= 4 \cdot (x - 1)\left( x + \frac{1}{4} \right) =\]

\[= (x - 1)(4x + 1);\]

\[\Longrightarrow \frac{x + 4^{\backslash 4x + 1}}{x - 1} -\]

\[- \frac{37x - 12}{(x - 1)(4x + 1)} =\]

\[= \frac{4x^{2} + x + 16x + 4 - 37x + 12}{(x - 1)(4x + 1)} =\]

\[= \frac{4x^{2} - 20x + 16}{(x - 1)(4x + 1)} =\]

\[= \frac{4 \cdot \left( x^{2} - 5x + 4 \right)}{(x - 1)(4x + 1)};\]

\[x^{2} - 5x + 4 = 0\]

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

\[x_{1,2} = \frac{5 \pm 3}{2} = 4;1;\]

\[x^{2} - 5x + 4 = (x - 4)(x - 1);\]

\[\Longrightarrow \frac{4 \cdot \left( x^{2} - 5x + 4 \right)}{(x - 1)(4x + 1)} =\]

\[= \frac{4 \cdot (x - 4)(x - 1)}{(x - 1)(4x + 1)} = \frac{4x - 16}{4x + 1}.\]

\[\textbf{б)}\ \frac{x - 1}{x + 2} - \frac{1 - x}{x^{2} + 3x + 2} =\]

\[= \frac{x - 1}{x + 2} - \frac{1 - x}{(x + 1)(x + 2)} =\]

\[= \frac{x - 1^{\backslash x + 1}}{x + 2} + \frac{x - 1}{(x + 1)(x + 2)} =\]

\[= \frac{(x - 1)(x + 1) + (x - 1)}{(x + 1)(x + 2)} =\]

\[= \frac{(x - 1)(x + 1 + 1)}{(x + 1)(x + 2)} =\]

\[= \frac{(x - 1)(x + 2)}{(x + 1)(x + 2)} = \frac{x - 1}{x + 1}\]

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

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

\[x_{1} = - 2;\ \ x_{2} = - 1.\]

\[\textbf{в)}\ \frac{7x - x^{2}}{x + 4} \cdot \frac{x^{2} - x - 20}{7 - x}\]

\[x^{2} - x - 20 = 0\]

\[x_{1} + x_{2} = 1;\ \ \ \ x_{1} \cdot x_{2} = - 20\]

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

\[x^{2} - x - 20 = (x - 5)(x + 4).\]

\[\Longrightarrow \frac{7x - x^{2}}{x + 4} \cdot \frac{x^{2} - x - 20}{7 - x} =\]

\[= \frac{x(7 - x)}{x + 4} \cdot \frac{(x - 5)(x + 4)}{7 - x} =\]

\[= x^{2} - 5x.\]

\[\textbf{г)}\ \ \frac{x^{2} + 11x + 30}{3x - 15}\ :\frac{x + 5}{x - 5}\]

\[x^{2} + 11x + 30 = 0\]

\[x_{1} + x_{2} = - 11;\ \ x_{1} \cdot x_{2} = 30\]

\[x_{1} = - 6;\ \ \ x_{2} = - 5;\]

\[x^{2} + 11x + 30 = (x + 6)(x + 5).\]

\[\Longrightarrow \frac{x^{2} + 11x + 30}{3x - 15}\ :\frac{x + 5}{x - 5} =\]

\[= \frac{(x + 6)(x + 5)}{3 \cdot (x - 5)} \cdot \frac{(x - 5)}{(x + 5)} =\]

\[= \frac{x + 6}{3}.\]

\[\textbf{д)}\ \frac{2x^{2} - 7}{x^{2} - 3x - 4} - \frac{x + 1}{x - 4} =\]

\[= \frac{2x^{2} - 7}{(x - 4)(x + 1)} - \frac{x + 1^{\backslash x + 1}}{x - 4}\]

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

\[x_{1} + x_{2} = 3;\ \ x_{1} \cdot x_{2} = - 4\]

\[x_{1} = 4;\ \ \ x_{2} = - 1\]

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

\[\Longrightarrow \frac{2x^{2} - 7}{(x - 4)(x + 1)} -\]

\[- \frac{x^{2} + 2x + 1}{(x - 4)(x + 1)} = \frac{x + 2}{x + 1}.\]

\[\textbf{е)}\ \frac{2 + x - x^{2}}{2 - 5x + 3x^{2}} + \frac{10x}{3x - 2}\]

\[3x^{2} - 5x + 2 = 0\]

\[D = 25 - 4 \cdot 3 \cdot 2 = 1\]

\[x_{1} = \frac{5 + 1}{6} = 1;\ \ \ x_{2} = \frac{5 - 1}{6} =\]

\[= \frac{4}{6} = \frac{2}{3};\]

\[3x^{2} - 5x + 2 =\]

\[= 3 \cdot (x - 1)\left( x - \frac{2}{3} \right) =\]

\[= (x - 1)(3x - 2);\]

\[\Longrightarrow \frac{2 + x - x^{2}}{(x - 1)(3x - 2)} + \frac{10x}{3x - 2} =\]

\[= \frac{2 + x - x^{2} + 10x(x - 1)}{(x - 1)(3x - 2)} =\]

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

\[= \frac{(3x - 2)(3x - 1)}{(x - 1)(3x - 2)} = \frac{3x - 1}{x - 1}.\]

\[\boxed{\text{228.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[(x + 1)\left( x^{4} - 6x^{2} + 5 \right) = 0\]

\[1)\ x + 1 = 0,\ \ x_{1} = - 1;\]

\[2)\ x^{4} - 6x^{2} + 5 = 0\]

\[Пусть\ x^{2} = t,\ \ t \geq 0:\]

\[t^{2} - 6t + 5 = 0\]

\[D = 3^{2} - 5 = 4\]

\[t_{1,2} = 3 \pm 2 = 1;5;\]

\[\Longrightarrow \left\{ \begin{matrix} x² = 1 \\ x² = 5 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x_{2,3} = \pm 1\ \ \ \ \\ x_{4,5} = \pm \sqrt{5.} \\ \end{matrix} \right.\ \]

\[Ответ:x = \pm 1;\ \ x = \pm \sqrt{5}.\]

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

\[1)\ x - 1 = 0,\ \ x_{1} = 1;\]

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

\[Пусть\ x^{2} = t,\ \ \ t \geq 0:\]

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

\[D = 1 + 3 = 4\]

\[t_{1,2} = 1 \pm 2 = 3;\ - 1.\]

\[Так\ как\ t \geq 0,\ то\ t = 3 \Longrightarrow x^{2} =\]

\[= 3 \Longrightarrow x_{2,3} = \pm \sqrt{3}.\]

\[Ответ:x = 1;\ \ x = \pm \sqrt{3}.\]