ГДЗ по алгебре 9 класс Макарычев ФГОС Задание 216

\[\boxed{\text{217.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ y³ - 6y = 0\]

\[y\left( y^{2} - 6 \right) = 0\]

\[y_{1} = 0\ \ \ \ \ и\ \ \ y_{2}^{2} = 6\]

\[y_{1} = 0,\ \ y_{2} = \sqrt{6},\ \ \]

\[y_{3} = - \sqrt{6};\]

\[Ответ:y = 0;y = \pm \sqrt{6}.\]

\[\textbf{б)}\ 6x^{4} + 3,6x^{2} = 0\]

\[x^{2}\left( 6x^{2} + 3,6 \right) = 0\]

\[x_{1} = 0\ \ \ \ и\ \ \ 6x^{2} = - 3,6 \Longrightarrow\]

\[\Longrightarrow корней\ нет;\]

\[\Longrightarrow x = 0\]

\[Ответ:x = 0.\]

\[\textbf{в)}\ x³ + 3x = 3,5x²\]

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

\[x\left( x^{2} - 3,5x + 3 \right) = 0\]

\[x_{1} = 0\ \ \ и\ \ x^{2} - 3,5x + 3 = 0\]

\[D = 12,25 - 4 \cdot 3 = 0,25\]

\[x_{2,3} = \frac{3,5 \pm 0,5}{2} = 2;1,5.\]

\[Ответ:x = 0;x = 2;x = 1,5.\]

\[\textbf{г)}\ x³ - 0,1x = 0,3x²\]

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

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

\[x_{1} = 0\ \ \ \ и\ \ \ \ x^{2} - 0,3x - 0,1 = 0\]

\[D = {0,3}^{2} + 4 \cdot 0,1 =\]

\[= 0,09 + 0,4 = 0,49\]

\[x_{2,3} = \frac{0,3 \pm 0,7}{2} \Longrightarrow x_{2} = 0,5;\ \]

\[\text{\ \ \ \ }x_{3} = - 0,2;\]

\[Ответ:x = 0;x = 0,5;x = - 0,2.\]

\[\textbf{д)}\ 9x³ - 18x^{2} - x + 2 = 0\]

\[9x^{2}(x - 2) - (x - 2) = 0\]

\[\left( 9x^{2} - 1 \right)(x - 2) = 0\]

\[(3x - 1)(3x + 1)(x - 2) = 0\]

\[x_{1} = \frac{1}{3},\ \ x_{2} = - \frac{1}{3},\]

\[\text{\ \ }x_{3} = 2\]

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

\[\textbf{е)}\ y^{4} - y^{3} - 16y^{2} + 16y = 0\]

\[y^{3}(y - 1) - 16y(y - 1) = 0\]

\[\left( y^{3} - 16y \right)(y - 1) = 0\]

\[y\left( y^{2} - 16 \right)(y - 1) = 0\]

\[y(y - 4)(y + 4)(y - 1) = 0\]

\[y_{1} = 0,\ \ y_{2} = 4,\ \ y_{3} = - 4,\]

\[\text{\ \ }y_{4} = 1;\]

\[Ответ:y = 0;y = 1;y = \pm 4.\]

\[\textbf{ж)}\ p³ - p^{2} = p - 1\]

\[p^{2}(p - 1) - (p - 1) = 0\]

\[\left( p^{2} - 1 \right)(p - 1) = 0\]

\[p_{1} = - 1,\ \ p_{2} = 1;\]

\[Ответ:p = \pm 1.\]

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

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

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

\[x(x - 3)(x - 1)(x + 1) = 0\]

\[x_{1} = 0,\ \ x_{2} = 3,\ \ \]

\[x_{3} = 1,\ \ x_{4} = - 1.\]

\[Ответ:x = 0;x = 3;x = \pm 1.\]