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МатематикаРешебник по алгебре 8 класс Макарычев ФГОС Задание 607
Задание 607
\[\boxed{\text{607\ (607).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[y \neq 0\]
\[y - 2 \neq 0,\ \ y \neq 2\]
\[y - 3 \neq 0,\ \ y \neq 3\]
\[5y \cdot (y - 3) - 4y \cdot (y - 2) =\]
\[= (y - 2)(y - 3)\]
\[5y^{2} - 15y - 4y^{2} + 8y =\]
\[= y^{2} - 3y - 2y + 6\]
\[- 2y = 6\]
\[y = - 3\]
\[Ответ:y = - 3.\]
\[x + 1 \neq 0,\ \ x \neq - 1\]
\[x + 2 \neq 0,\ \ x \neq - 2\]
\[x + 3 \neq 0,\ \ x \neq - 3\ \]
\[- 3x^{2} - 5x = 0\]
\[- x(3x + 5) = 0\]
\[x = 0\ \ \ \ \ \ \ \ \ \ \ \ 3x + 5 = 0\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = - \frac{5}{3}\]
\[Ответ:x = - 1\frac{2}{3};\ \ x = 0.\]
\[\textbf{в)}\ \frac{1}{x + 2} + \frac{1}{x^{2} - 2x} = \frac{8}{x^{3} - 4x}\ \]
\[x \neq 0\]
\[x - 2 \neq 0,\ \ x \neq 2\]
\[x + 2 \neq 0,\ \ x \neq - 2\]
\[x(x - 2) + x + 2 = 8\]
\[x^{2} - 2x + x + 2 = 8\]
\[x^{2} - x - 6 = 0\]
\[D = 1 + 24 = 25\]
\[x_{1,2} = \frac{1 \pm 5}{2}\]
\[x_{1} = 3;\ \ x_{2} = - 2 -\]
\[не\ принадлежит\ по\ ОДЗ\]
\[Ответ:x = 3.\]
\[\textbf{г)}\frac{10}{y^{3} - y} + \frac{1}{y - y^{2}} = \frac{1}{1 + y}\]
\[\frac{10}{y(y - 1)(y + 1)} + \frac{1}{y(1 - y)} =\]
\[= \frac{1}{1 + y}\]
\[y \neq 0\]
\[y - 1 \neq 0,\ \ y \neq 1\]
\[y + 1 \neq 0,\ \ y \neq - 1\]
\[10 - (y + 1) = y(y - 1)\]
\[10 - y - 1 = y^{2} - y\]
\[y^{2} = 9\]
\[y = \pm 3\]
\[Ответ:y = \pm 3.\]
\[\textbf{д)}\ 1 + \frac{45}{x^{2} - 8x + 16} = \frac{14}{x - 4}\]
\[x - 4 \neq 0,\ \ x \neq 4\]
\[(x - 4)^{2} + 45 = 14 \cdot (x - 4)\]
\[x^{2} - 8x + 16 + 45 = 14x - 56\]
\[x^{2} - 22x + 117 = 0\]
\[D = 484 - 468 = 16\]
\[x_{1,2} = \frac{22 \pm 4}{2}\]
\[x_{1} = 13;\ \ x_{2} = 9\]
\[Ответ:x = \left\{ 9;13 \right\}.\]
\[\textbf{е)}\frac{5}{x - 1} - \frac{4}{3 - 6x + 3x^{2}} = 3\]
\[x - 1 \neq 0,\ \ x \neq 1\]
\[\frac{5}{x - 1} - \frac{4}{3 \cdot \left( 1 - 2x + x^{2} \right)} = 3\]
\[\frac{5}{x - 1} - \frac{4}{3 \cdot (1 - x)^{2}} = 3\]
\[15 \cdot (x - 1) - 4 = 9 \cdot (x - 1)^{2}\]
\[15x - 15 - 4 =\]
\[= 9 \cdot \left( x^{2} - 2x + 1 \right)\]
\[15x - 19 = 9x^{2} - 18x + 9\]
\[9x^{2} - 33x + 28 = 0\]
\[D = 1089 - 1008 = 81\]
\[x_{1,2} = \frac{33 \pm 9}{18} = \frac{42}{18};\ \frac{24}{18}\]
\[x_{1} = \frac{7}{3};\ \ x_{2} = \frac{4}{3}\]
\[Ответ:x = 1\frac{1}{3};\ \ x = 2\frac{1}{3}\text{.\ }\]
\[\boxed{\text{607.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]
Пояснение.
Формулы квадрата суммы и квадрата разности:
\[a^{2} - 2ab + b^{2} = (a - b)^{2};\]
\[a^{2} + 2ab + b^{2} = (a + b)^{2}.\]
Решение.
\[\textbf{а)}\ x² - 6x - 2 =\]
\[= \left( x^{2} - 6x + 9 \right) - 11 =\]
\[= (x - 3)^{2} - 11;\]
\[\textbf{б)}\ x² + 5x + 20 =\]
\[= x^{2} + 5x + {2,5}^{2} - {2,5}^{2} + 20 =\]
\[= (x + 2,5)^{2} + 13,75;\]
\[\textbf{в)}\ 2x² - 4x + 10 =\]
\[= 2 \cdot \left( x^{2} - 2x + 5 \right) =\]
\[= 2 \cdot \left( x^{2} - 2x + 1 + 4 \right) =\]
\[= 2 \cdot (x - 1)^{2} + 8;\]
\[\textbf{г)}\ \frac{1}{2}x² + x - 6 =\]
\[= \frac{1}{2} \cdot \left( x^{2} + 2x - 12 \right) =\]
\[= \frac{1}{2} \cdot \left( x^{2} + 2x + 1 - 13 \right) =\]
\[= \frac{1}{2} \cdot (x + 1)^{2} - 6,5.\]
