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МатематикаРешебник по алгебре 9 класс Макарычев Задание 59
Задание 59
\[\boxed{\text{59\ (59).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[\textbf{а)}\ x^{2} + x - 6 = 0\]
\[D = 1 + 6 \cdot 4 = 25 = 5^{2}\]
\[x_{1} = \frac{- 1 + 5}{2} = 2;\ \ \ x_{2} =\]
\[= \frac{- 1 - 5}{2} = - 3.\]
\[Ответ:x = 2;\ \ x = - 3.\]
\[\textbf{б)}\ 9x^{2} - 9x + 2 = 0\]
\[D = 9^{2} - 4 \cdot 9 \cdot 2 = 81 - 72 = 9\]
\[x_{1} = \frac{9 + 3}{18} = \frac{12}{18} = \frac{2}{3};\ \ x_{2} =\]
\[= \frac{9 - 3}{18} = \frac{6}{18} = \frac{1}{3}.\]
\[Ответ:x = \frac{1}{3};\ \ x = \frac{2}{3}.\]
\[\textbf{в)}\ 0,2x^{2} + 3x - 20 = 0\ | \cdot 5\]
\[x^{2} + 15x - 100 = 0\]
\[D = 15^{2} + 400 = 225 + 400 =\]
\[= 625\]
\[x_{1} = \frac{- 15 - 25}{2} = - \frac{40}{2} = - 20;\ \ \ \]
\[x_{2} = \frac{- 15 + 25}{2} = - 5.\]
\[Ответ:x = - 20;\ \ x = 5.\]
\[\textbf{г)} - 2x^{2} - x - 0,125 = 0\ | \cdot ( - 8)\]
\[16x^{2} + 8x + 1 = 0\]
\[(4x + 1)^{2} = 0\]
\[4x + 1 = 0\]
\[4x = - 1\]
\[x = - \frac{1}{4} = - 0,25\]
\[Ответ:x = - 0,25.\]
\[\textbf{д)}\ 0,1x^{2} + 0,4 = 0\]
\[0,1x^{2} = - 0,4\]
\[x^{2} = - 4 \Longrightarrow корней\ нет.\]
\[Ответ:решений\ нет.\]
\[\textbf{е)} - 0,3x^{2} + 1,5x = 0\ |\ :( - 0,3)\]
\[x^{2} - 5x = 0\]
\[x(x - 5) = 0\]
\[x_{1} = 0\ \ и\ \ \ x_{2} = 5.\]
\[Ответ:x = 0;x = 5.\]
\[\boxed{\text{59.\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[1)\ \frac{2ab}{a^{2} - b^{2}} + \frac{a - b}{2a + 2b} =\]
\[= \frac{2ab^{\backslash 2}}{(a - b)(a + b)} + \frac{a - b^{\backslash a - b}}{2 \cdot (a + b)} =\]
\[= \frac{4ab + a^{2} - 2ab + b^{2}}{2 \cdot (a - b)(a + b)} =\]
\[= \frac{a^{2} + 2ab + b^{2}}{2 \cdot (a - b)(a + b)} =\]
\[= \frac{(a + b)^{2}}{2 \cdot (a - b)(a + b)} = \frac{a + b}{2(a - b)}\]
\[2)\ \frac{a + b}{2(a - b)} \cdot \frac{2a}{a + b} = \frac{a}{a - b}\]
\[3)\ \frac{a}{a - b} - \frac{b}{a - b} = \frac{a - b}{a - b} = 1\]
\[1)\ \frac{x}{(x - y)^{2}} - \frac{y}{x^{2} - y^{2}} =\]
\[= \frac{x^{\backslash x + y}}{(x - y)(x - y)} - \frac{y^{\backslash x - y}}{(x - y)(x + y)} =\]
\[= \frac{x^{2} + xy - xy + y^{2}}{(x - y)(x - y)(x + y)} =\]
\[= \frac{x^{2} + y^{2}}{(x^{2} - y^{2})(x - y)}\]
\[2)\ \frac{x\left( x^{2} - y^{2} \right)}{x^{2} + y^{2}} \cdot \frac{x^{2} + y^{2}}{\left( x^{2} - y^{2} \right)(x - y)} =\]
\[= \frac{x}{x - y}\]
\[3)\ \frac{y}{x - y} - \frac{x}{x - y} = \frac{y - x}{x - y} =\]
\[= - \frac{x - y}{x - y} = - 1\]
