ГДЗ по алгебре 9 класс Макарычев Задание 379

 

\[\boxed{\text{379\ (379).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[(a + 2)x^{2} + 8x + a - 4 = 0\]

\[Квадратное\ уравнение\ имеет\ \]

\[два\ корня,\ если\ D > 0.\]

\[D = 16 - (a + 2)(a - 4) =\]

\[= 16 - a^{2} + 2a + 8 =\]

\[= 24 + 2a - a^{2}\]

\[D > 0:\]

\[a^{2} - 2a - 24 < 0\]

\[D_{1} = 1 + 24 = 25\]

\[a_{1} = 1 + 5 = 6;\ \ a_{2} =\]

\[= 1 - 5 = - 4.\]

\[(a + 4)(a - 6) < 0\]

\[a + 2 \neq 0\ \ \]

\[a \neq - 2.\]

\[при\ a \in ( - 4; - 2) \cup ( - 2;6).\]

\[\boxed{\text{379.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ x^{2} + y^{2} = 4;\]

\[x^{2} + y^{2} = 25.\]

\[\textbf{б)}\ (x - 3)^{2} + y^{2} = 4;\]

\[(x - 3)^{2} + y^{2} = 25.\]

\[\textbf{в)}\ x^{2} + (y - 4)^{2} = 4;\]

\[x^{2} + (y - 4)^{2} = 25.\]

\[\textbf{г)}\ (x + 1)^{2} + (y - 2)^{2} = 4;\]

\[(x + 1)^{2} + (y - 2)^{2} = 25.\]