ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 608

 

\[\boxed{\text{608\ (608).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[x - 5 \neq 0,\ \ x \neq 5\]

\[x + 1 \neq 0,\ \ x \neq - 1\]

\[10 + x(x - 5) = 3(x + 1)\]

\[10 + x^{2} - 5x = 3x + 3\]

\[x^{2} - 8x + 7 = 0\]

\[D = 64 - 28 = 36\]

\[x_{1,2} = \frac{8 \pm 6}{2}\]

\[x_{1} = 7;\ \ x_{2} = 1\]

\[Ответ:x = \left\{ 1;7 \right\}.\]

\[x - 3 \neq 0,\ \ x \neq 3\]

\[x + 4 \neq 0,\ \ x \neq - 4\]

\[17 - (x + 4) = x(x - 3)\]

\[17 - x - 4 = x^{2} - 3x\]

\[x^{2} - 2x - 13 = 0\]

\[D = 4 + 52 = 56 = 4 \cdot 14\]

\[Ответ:x = \left\{ 1 - \sqrt{14};1 + \sqrt{14} \right\}.\]

\[x^{2} \neq 1,\ \ x \neq \pm 1\]

\[4x^{2} - 10x + 2 = 0\ \ \ \ \ |\ :2\]

\[2x^{2} - 5x + 1 = 0\]

\[D = 25 - 8 = 17\]

\[x_{1,2} = \frac{5 \pm \sqrt{17}}{4}\]

\[Ответ:x = \left\{ \frac{5 - \sqrt{17}}{4};\frac{5 + \sqrt{17}}{4} \right\}.\]

\[\textbf{г)}\frac{4}{9x^{2} - 1} + \frac{1}{3x^{2} - x} =\]

\[= \frac{4}{9x^{2} - 6x + 1}\]

\[x \neq 0\]

\[3x + 1 \neq 0,\ \ x \neq - \frac{1}{3}\]

\[3x - 1 \neq 0,\ \ x \neq \frac{1}{3}\]

\[9x^{2} - 8x - 1 = 0\]

\[D = 64 + 36 = 100\]

\[x_{1,2} = \frac{8 \pm 10}{18}\]

\[x_{1} = 1;\ \ x_{2} = - \frac{1}{9}\]

\[Ответ:x = \left\{ - \frac{1}{9};1 \right\}\text{.\ }\]

\[\boxed{\text{608.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

Пояснение.

Формулы квадрата суммы и квадрата разности:

\[a^{2} - 2ab + b^{2} = (a - b)^{2};\]

\[a^{2} + 2ab + b^{2} = (a + b)^{2}.\]

Решение.

\[\textbf{а)}\ x² - 10x + 10 =\]

\[= x^{2} - 10x + 25 - 25 + 10 =\]

\[= \left( x^{2} - 10x + 25 \right) - 15 =\]

\[= (x - 5)^{2} - 15;\]

\[\textbf{б)}\ x² + 3x - 1 =\]

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

\[= \left( x + \frac{3}{2} \right)^{2} - \frac{9}{4} - 1 =\]

\[= (x + 1,5)^{2} - 3,25;\]

\[\textbf{в)}\ 3x² + 6x - 3 =\]

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

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

\[= 3 \cdot \left( x^{2} + 2x + 1 \right) - 6 =\]

\[= 3 \cdot (x + 1)^{2} - 6;\]

\[\textbf{г)}\frac{1}{4}x² - x + 2 =\]

\[= \frac{1}{4} \cdot \left( x^{2} - 4x + 8 \right) =\]

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

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