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МатематикаГДЗ по алгебре 8 класс Макарычев ФГОС Задание 546
Задание 546
\[\boxed{\text{546\ (546).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
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\[\textbf{а)}\ \frac{x^{2} - 1}{2} - 11x = 11\ \ \ \ \ | \cdot 2\] \[x^{2} - 1 - 22x = 22\] \[x^{2} - 22x - 23 = 0\] \[D_{1} = 121 + 23 = 144\] \[x_{1,2} = 11 \pm \sqrt{144} = 11 \pm 12.\] \[x_{1} = 23;\ \ x_{2} = - 1.\] |
\[\textbf{б)}\ \frac{x^{2} + x}{2} = \frac{8x - 7}{3}\ \ \ \ \ \ | \cdot 6\] \[3x^{2} + 3x = 16x - 14\] \[3x^{2} + 3x - 16x + 14 = 0\] \[3x^{2} - 13x + 14 = 0\] \[D = 169 - 168 = 1\] \[x_{1,2} = \frac{13 \pm \sqrt{1}}{2 \cdot 3} = \frac{13 \pm 1}{6}\] \[x_{1} = 2;\ \ x_{2} = \frac{14}{6} = 2\frac{1}{3}.\] |
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\[\textbf{в)}\ \frac{4x^{2} - 1}{3} = x(10x - 9)\ \ \ \ | \cdot 3\] \[4x^{2} - 1 = 30x^{2} - 27x\] \[- 26x^{2} + 27x - 1 = 0\] \[26x^{2} - 27x + 1 = 0\] \[D = 729 - 104 = 625\] \[x_{1,2} = \frac{- 27 \pm \sqrt{625}}{- 26 \cdot 2} = \frac{- 27 \pm 25}{- 52}\] \[x_{1} = \frac{1}{26};\ \ \ x_{2} = 1.\] |
\[\textbf{г)}\ \frac{3}{4}x^{2} - \frac{2}{5}x = \frac{4}{5}x^{2} + \frac{3}{4}\ \ \ \ \ \ | \cdot 20\] \[15x^{2} - 8x = 16x^{2} + 15\] \[15x^{2} - 8x - 16x^{2} - 15 = 0\] \[- x^{2} - 8x - 15 = 0\] \[x^{2} + 8x + 15 = 0\ \] \[D_{1} = 4^{2} - 15 = 16 - 15 = 1\] \[x_{1,2} = - 4 \pm \sqrt{1} = - 4 \pm 1\] \[x_{1} = - 5;\ \ x_{2} = - 3.\ \ \] |
\[\boxed{\text{546.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]
Пояснение.
Чтобы решить уравнение графически, нужно построить два графика функции в одной координатной плоскости и найти точки их пересечения.
y=x² - это один график, остальные данные переносим вправо, меняя знаки на противоположные, приравниваем к y и получаем второй график.
Решение.
\[\textbf{а)}\ x^{2} - 2x - 1 = 0\]
\[x^{2} = 2x + 1\]
\[\left\{ \begin{matrix} y = x^{2}\text{\ \ \ \ \ \ \ } \\ y = 2x + 1 \\ \end{matrix} \right.\ \]
\[y = x^{2}\]
| \[x\] | \[0\] | \[1\] | \[2\] | \[- 1\] | \[- 2\] |
|---|---|---|---|---|---|
| \[y\] | \[0\] | \[1\] | \[4\] | \[1\] | \[4\] |
\[y = 2x + 1\]
| \[x\] | \[0\] | \[1\] |
|---|---|---|
| \[y\] | \[1\] | \[3\] |
\[x^{2} - 2x - 1 = 0\]
\[D = 4 + 4 = 8\]
\[x_{1} = \frac{2 - \sqrt{8}}{2} = \frac{2 - 2\sqrt{2}}{2} = - 0,4\]
\[x_{2} = \frac{2 + \sqrt{8}}{2} = \frac{2 + 2\sqrt{2}}{2} = 2,4\]
\[Ответ:x = - 0,4;\ \ x = 2,4.\ \]
\[\textbf{б)}\ x^{2} - 4x + 2 = 0\]
\[x^{2} = 4x - 2\]
\[\left\{ \begin{matrix} y = x^{2} \\ y = 4x - 2 \\ \end{matrix} \right.\ \]
\[y = x^{2}\]
| \[x\] | \[0\] | \[1\] | \[2\] | \[- 1\] | \[- 2\] |
|---|---|---|---|---|---|
| \[y\] | \[0\] | \[1\] | \[4\] | \[1\] | \[4\] |
\[y = 4x - 2\]
| \[x\] | \[0\] | \[1\] |
|---|---|---|
| \[y\] | \[- 2\] | \[2\] |
\[x^{2} - 4x + 2 = 0\]
\[D = 16 - 8 = 8\]
\[x_{1} = \frac{4 - \sqrt{8}}{2} = \frac{4 - 2\sqrt{2}}{2} = 0,6\]
\[x_{2} = \frac{4 + \sqrt{8}}{2} = \frac{4 + 2\sqrt{2}}{2} = 3,4\]
\(\ \)
\[Ответ:x = 3,4;\ \ x = 0,6.\]
