Решебник по алгебре 8 класс Макарычев ФГОС Задание 609

 

\[\boxed{\text{609\ (609).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

\[x - 2 \neq 0,\ \ x \neq 2\]

\[x \neq 0\]

\[11x^{2} - 64x - 12 = 0\]

\[D = 4096 + 528 = 4624 = 68^{2}\]

\[x_{1,2} = \frac{64 \pm 68}{22} = \frac{132}{22};\ - \frac{4}{22}\]

\[x_{1} = \frac{66}{11} = 6;\ \ x_{2} = - \frac{2}{11}\]

\[Ответ:x = \left\{ - \frac{2}{11};6 \right\}.\]

\[\textbf{б)}\frac{2}{y^{2} - 3y} - \frac{1}{y - 3} = \frac{5}{y^{3} - 9y}\]

\[y \neq 0\]

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

\[y + 3 \neq 0,\ \ y \neq - 3\]

\[2 \cdot (y + 3) - y \cdot (y + 3) = 5\]

\[2y + 6 - y^{2} - 3y = 5\]

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

\[D = 1 + 4 = 5\]

\[y_{1,2} = \frac{- 1 \pm \sqrt{5}}{2}\]

\[Ответ:y = \left\{ \frac{- 1 - \sqrt{5}}{2};\frac{- 1 + \sqrt{5}}{2} \right\}.\]

\[\textbf{в)}\frac{18}{4x^{2} + 4x + 1} - \frac{1}{2x^{2} - x} =\]

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

\[x \neq 0\]

\[2x - 1 \neq 0,\ \ x \neq \frac{1}{2}\]

\[2x + 1 \neq 0,\ \ x \neq - \frac{1}{2}\]

\[18x(2x - 1) - (2x + 1)^{2} =\]

\[= 6x(2x + 1)\]

\[36x^{2} - 18x - 4x^{2} - 4x - 1 =\]

\[= 12x^{2} + 6x\]

\[20x^{2} - 28x - 1 = 0\]

\[D = 784 + 80 = 864 = 144 \cdot 6\]

\[x_{1,2} = \frac{28 \pm \sqrt{144 \cdot 6}}{40} =\]

\[= \frac{28 \pm 12\sqrt{6}}{40} = \frac{7 \pm 3\sqrt{6}}{10}\]

\[Ответ:x = \left\{ \frac{7 - 3\sqrt{6}}{10};\frac{7 + 3\sqrt{6}}{10} \right\}.\]

\[\textbf{г)}\frac{3 \cdot \left( 4y^{2} + 10y - 7 \right)}{16y^{2} - 9} =\]

\[= \frac{3y - 7}{3 - 4y} + \frac{6y + 5}{3 + 4y}\]

\[4y - 3 \neq 0,\ \ y \neq \frac{3}{4}\]

\[4y + 3 \neq 0,\ \ y \neq - \frac{3}{4}\]

\[9y = 27\]

\[y = 3\]

\[Ответ:y = 3\text{.\ }\]

\[\boxed{\text{609.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

Пояснение.

Формулы квадрата суммы и квадрата разности:

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

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

Решение.

\[\textbf{а)}\ x² - 6x + 10 =\]

\[= \left( x^{2} - 6x + 9 \right) + 1 =\]

\[= (x - 3)^{2} + 1 > 0;\]

\[\textbf{б)}\ 5x² - 10x + 5 =\]

\[= 5 \cdot \left( x^{2} - 2x + 1 \right) =\]

\[= 5 \cdot (x - 1)^{2} > 0;\]

\[\textbf{в)} - x^{2} + 20x - 100 =\]

\[= - \left( x^{2} - 20x + 100 \right) =\]

\[= - (x - 10)^{2} \leq 0;\]

\[\textbf{г)} - 2x^{2} + 16x - 33 =\]

\[= - \left( 2x^{2} - 16x + 33 \right) =\]

\[= - \left( x^{2} - 8x + 16 \right) \cdot 2 - 1 =\]

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

\[= - \left( 2 \cdot (x - 4)^{2} + 1 \right) < 0.\]

\[\textbf{д)}\ x^{2} - 0,32x + 0,0256 =\]

\[= (x - 0,16)^{2} \geq 0.\]

\[\textbf{е)}\ 4x^{2} + 0,8x + 2 =\]

\[= 4 \cdot \left( x^{2} + 0,2x + 0,5 \right) =\]

\[= 4 \cdot \left( x^{2} + 0,2x + 0,01 \right) + 0,49 =\]

\[= 4 \cdot (x + 0,1)^{2} + 0,49 > 0.\]