ГДЗ по алгебре 7 класс Макарычев ФГОС Задание 1029

\[\boxed{\text{1029.}\text{\ }еуроки - ответы\ на\ пятёрку}\]

\[\textbf{а)}\ x³ + 3x² - 4x - 12 = 0\]

\[x^{2}(x + 3) - 4 \cdot (x + 3) = 0\]

\[(x + 3)\left( x^{2} - 4 \right) = 0\]

\[x + 3 = 0\ \ \ \ \ или\ \ \ \ \ \]

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

\[x = - 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \]

\[x = 2\ \ \ \ или\mathbf{\text{\ \ \ }}\ \ x = - 2\]

\[Ответ:x = - 3;x = \pm 2.\]

\[\textbf{б)}\ 2m³ - m^{2} - 18m + 9 = 0\]

\[m^{2}(2m - 1) - 9 \cdot (2m - 1) = 0\]

\[(2m - 1)\left( m^{2} - 9 \right) = 0\]

\[(2m - 1)(m - 3)(m + 3) = 0\]

\[2m = 1\ \ \ \ \ или\ \ \ \ m = 3\ \ \ \ \]

\[или\ \ \ m = - 3\]

\[m = 0,5\]

\[Ответ:m = 0,5;\ \ m = \pm 3.\]

\[\textbf{в)}\ y³ - 6y^{2} = 6 - y\]

\[y^{2}(y - 6) - 6 + y = 0\]

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

\[y = 6\ \ \ или\ \ \ y^{2} =\]

=\(- 1 \Longrightarrow неверно,\ при\ \)

\[любом\ значении\ y.\]

\[Ответ:y = 6.\]

\[\textbf{г)}\ 2a³ + 3a² = 2a + 3\]

\[a^{2}(2a + 3) - (2a + 3) = 0\]

\[(2a + 3)\left( a^{2} - 1 \right) = 0\]

\[(2a + 3)(a - 1)(a + 1) = 0\]

\[2a = - 3\ \ \ \ или\ \ \ \ a = 1\ \ \ \ \ или\ \ \]

\[\ a = - 1\]

\[a = - 1,5.\]

\[Ответ:a = - 1,5;\ \ a = \pm 1.\]