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МатематикаРешебник по алгебре 9 класс Макарычев Задание 272
Задание 272
\[\boxed{\text{272}\text{\ (272)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ y^{3} - 6y = 0\]
\[y\left( y^{2} - 6 \right) = 0\]
\[y_{1} = 0\ \ \ \ \ и\ \ \ y_{2}^{2} = 6\]
\[y_{1} = 0,\ \ y_{2} = \sqrt{6},\ \ \]
\[y_{3} = - \sqrt{6};\]
\[Ответ:y = 0;y = \pm \sqrt{6}.\]
\[\textbf{б)}\ 6x^{4} + 3,6x^{2} = 0\]
\[x^{2}\left( 6x^{2} + 3,6 \right) = 0\]
\[x_{1} = 0\ \ \ \ и\ \ \ 6x^{2} = - 3,6 \Longrightarrow\]
\[\Longrightarrow корней\ нет;\]
\[\Longrightarrow x = 0\]
\[Ответ:x = 0.\]
\[\textbf{в)}\ x^{3} + 3x = 3,5x^{2}\]
\[x^{3} - 3,5x^{2} + 3x = 0\]
\[x\left( x^{2} - 3,5x + 3 \right) = 0\]
\[x_{1} = 0\ \ \ и\ \ x^{2} - 3,5x + 3 = 0\]
\[D = 12,25 - 4 \cdot 3 = 0,25\]
\[x_{2,3} = \frac{3,5 \pm 0,5}{2} = 2;1,5.\]
\[Ответ:x = 0;x = 2;x = 1,5.\]
\[\textbf{г)}\ x^{3} - 0,1x = 0,3x^{2}\]
\[x^{3} - 0,3x^{2} - 0,1x = 0\]
\[x\left( x^{2} - 0,3x - 0,1 \right) = 0\]
\[x_{1} = 0\ \ \ \ и\ \ \ \ x^{2} - 0,3x - 0,1 = 0\]
\[D = {0,3}^{2} + 4 \cdot 0,1 = 0,09 +\]
\[+ 0,4 = 0,49\]
\[x_{2,3} = \frac{0,3 \pm 0,7}{2} \Longrightarrow x_{2} = 0,5;\ \ \ \ \]
\[\ x_{3} = - 0,2;\]
\[Ответ:x = 0;x = 0,5;x = - 0,2.\]
\[\textbf{д)}\ 9x^{3} - 18x^{2} - x + 2 = 0\]
\[9x^{2}(x - 2) - (x - 2) = 0\]
\[\left( 9x^{2} - 1 \right)(x - 2) = 0\]
\[(3x - 1)(3x + 1)(x - 2) = 0\]
\[x_{1} = \frac{1}{3},\ \ x_{2} = - \frac{1}{3},\ \ \]
\[x_{3} = 2\]
\[Ответ:x = 2;\ \ x = \pm \frac{1}{3}.\]
\[\textbf{е)}\ y^{4} - y^{3} - 16y^{2} + 16y = 0\]
\[y^{3}(y - 1) - 16y(y - 1) = 0\]
\[\left( y^{3} - 16y \right)(y - 1) = 0\]
\[y\left( y^{2} - 16 \right)(y - 1) = 0\]
\[y(y - 4)(y + 4)(y - 1) = 0\]
\[y_{1} = 0,\ \ y_{2} = 4,\ \ y_{3} = - 4,\ \ \]
\[y_{4} = 1;\]
\[Ответ:y = 0;y = 1;y = \pm 4.\]
\[\textbf{ж)}\ p^{3} - p^{2} = p - 1\]
\[p^{2}(p - 1) - (p - 1) = 0\]
\[\left( p^{2} - 1 \right)(p - 1) = 0\]
\[p_{1} = - 1,\ \ p_{2} = 1;\]
\[Ответ:p = \pm 1.\]
\[\textbf{з)}\ x^{4} - x^{2} = 3x^{3} - 3x\]
\[x^{2}\left( x^{2} - 1 \right) - 3x\left( x^{2} - 1 \right) = 0\]
\[\left( x^{2} - 3x \right)\left( x^{2} - 1 \right) = 0\]
\[x(x - 3)(x - 1)(x + 1) = 0\]
\[x_{1} = 0,\ \ x_{2} = 3,\ \ \]
\[x_{3} = 1,\ \ x_{4} = - 1.\]
\[Ответ:x = 0;x = 3;x = \pm 1.\]
\[\boxed{\text{272.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ 3x² + 40x + 10 <\]
\[< - x^{2} + 11x + 3\]
\[4x^{2} + 29x + 7 < 0\]
\[D = 29^{2} - 4 \cdot 4 \cdot 7 = 729\]
\[x_{1,2} = \frac{- 29 \pm 27}{8},\ \ x_{1} = - \frac{1}{4},\]
\[\text{\ \ }x_{2} = - 7,\]
\[4 \cdot (x + 0,25)(x + 7) < 0\]
\[x \in ( - 7;\ - 0,25).\]
\[\textbf{б)}\ 9x² - x + 9 \geq 3x^{2} + 18x - 6\]
\[6x^{2} - 19x + 15 \geq 0\]
\[D = 19^{2} - 4 \cdot 6 \cdot 15 = 1\]
\[x_{1,2} = \frac{19 \pm 1}{12} = 1,5;1\frac{2}{3};\]
\[6 \cdot (x - 1,5)\left( x - 1\frac{2}{3} \right) \geq 0\]
\[x \in ( - \infty;1,5) \cup \left( 1\frac{2}{3};\ + \infty \right).\]
\[\textbf{в)}\ 2x² + 8x - 111 <\]
\[< (3x - 5)(2x + 6)\]
\[2x^{2} + 8x - 111 <\]
\[< 6x^{2} + 18x - 10x - 30\]
\[4x^{2} + 81 > 0 \Longrightarrow\]
\[\Longrightarrow x - любое\ число.\]
\[\textbf{г)}\ (5x + 1)(3x - 1) >\]
\[> (4x - 1)(x + 2)\]
\[15x^{2} - 5x + 3x - 1 >\]
\[> 4x^{2} + 8x - x - 2\]
\[11x^{2} - 9x + 1 > 0\]
\[D = 81 - 44 = 37\]
\[x_{1,2} = \frac{9 \pm \sqrt{37}}{22}\]
