Страница 61 номер 249, ГДЗ по алгебре за 9 класс к учебнику Мерзляка
Разложите на множители квадратный трёхчлен: 1) \(x^2 - x - 12\); 2) \(-x^2 + 2x + 35\); 3) \(6x^2 + 11x - 2\); 4) \(\frac{2}{3}x^2 + 3x - 6\).
1) \(x^{2} - x - 12 = x^{2} + 3x - 4x - 12 =\)
\(= x(x + 3) - 4(x + 3) = (x - 4)(x + 3)\)
2) \(-x^{2} + 2x + 35 = - \left( x^{2} - 2x - 35 \right) =\)
\(= - \left( x^{2} + 5x - 7x - 35 \right) = - \left( x(x + 5) - 7(x + 5) \right) =\)
\(= - (x + 5)(x - 7)\)
3) \(6x^{2} + 11x - 2\)
\(D = 11^{2} + 4 \cdot 6 \cdot 2 = 121 + 48 = 169 = 13^{2}\)
\(x_{1} = \dfrac{- 11 - 13}{12} = - 2;\ x_{2} = \dfrac{- 11 + 13}{12} = \dfrac{2}{12} = \dfrac{1}{6}\)
\(6x^{2} + 11x - 2 = 6(x + 2)\left( x - \dfrac{1}{6} \right)\)
4) \(\frac{2}{3}x^{2} + 3x - 6 = \frac{1}{3}(2x^{2} + 9x - 18)\)
\(D = 81 + 4 \cdot 2 \cdot 18 = 81 + 144 = 225 = 15^{2}\)
\(x_{1} = \dfrac{- 9 - 15}{4} = - 6;\ x_{2} = \dfrac{- 9 + 15}{4} = \dfrac{6}{4} = \dfrac{3}{2}\)
\(\dfrac{1}{3}\left( 2x^{2} + 9x - 18 \right) = \dfrac{1}{3} \cdot 2(x + 6)\left( x - \dfrac{3}{2} \right)\)
Ответ: 1) \((x - 4)(x + 3)\); 2) \(- (x + 5)(x - 7)\); 3) \(6(x + 2)\left( x - \frac{1}{6} \right)\); 4) \(\frac{2}{3}(x + 6)\left( x - \frac{3}{2} \right)\).
