ГДЗ по алгебре 8 класс Макарычев Задание 650

\[\boxed{\text{650\ (650).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ (x + 2)^{2} + (x - 3)^{2} = 13\]

\[x^{2} + 4x + 4 + x^{2} - 6x + 9 = 13\]

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

\[2x(x - 1) = 0\]

\[2x = 0\ \ \ \ \ \ \ \ x - 1 = 0\]

\[\ \ \ x = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 1\]

\[Ответ:x = \left\{ 0;1 \right\}.\]

\[\textbf{б)}\ (3x - 5)^{2} - (2x + 1)^{2} = 24\]

\[9x^{2} - 30x + 25 - 4x^{2} - 4x - 1 = 24\]

\[5x^{2} - 34x = 0\]

\[x(5x - 34) = 0\]

\[x = 0\ \ \ \ \ \ \ \ \ \ 5x = 34\ \ \ \]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 34\ :5\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 6,8\]

\[Ответ:x = \left\{ 0;6,8 \right\}.\]

\[\textbf{в)}\ (x - 4)\left( x^{2} + 4x + 16 \right) + 28 =\]

\[= x^{2}(x - 25)\]

\[25x^{2} = 36\]

\[x^{2} = \frac{36}{25}\]

\[x = \pm \frac{6}{5} = \pm 1,2\]

\[Ответ:x = \left\{ - 1,2;1,2 \right\}.\]

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

\[= 1,6x^{2}(5x - 2)\]

\[- 3,2x^{2} = 0\]

\[x = 0\]

\[Ответ:x = 0\text{.\ }\]