Решебник по алгебре 9 класс Макарычев Задание 105

 

\[\boxed{\text{105\ (105).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[(x + 3)^{2} - (x - 3)^{2} =\]

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

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

\[= x^{2} - 4x + 4 + x^{2} + 4x + 4\]

\[12x = 2x^{2} + 8\]

\[x^{2} - 6x + 4 = 0\]

\[D_{1} = 3^{2} - 4 = 9 - 4 = 5\]

\[x_{1,2} = 3 \pm \sqrt{5}\text{\ \ \ }\]

\[\boxed{\text{105.\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ f(x) = \frac{5}{x}\]

\[f( - x) = \frac{5}{- x} = - \frac{5}{x} = - f(x)\]

\[функция\ нечетная.\]

\[\textbf{б)}\ f(x) = 5 - 3x^{2}\]

\[f( - x) = 5 - 3x^{2} = f(x)\]

\[функция\ четная.\]

\[\textbf{в)}\ f(x) = x^{3} - x\]

\[f( - x) = - x^{3} + x =\]

\[= - \left( x^{3} - x \right) = - f(x)\]

\[функция\ нечетная.\]

\[\textbf{г)}\ f(x) = 1 - |x| = 1 - x\]

\[f( - x) = 1 - | - x| = 1 - x =\]

\[= f(x)\]

\[функция\ четная.\]