Страница 30 номер 108, ГДЗ по алгебре за 9 класс к учебнику Мерзляка
Решите уравнение: 1) \(9 - 7(x + 3) = 5 - 6x\); 2) \(\frac{x + 3}{2} - \frac{x - 4}{7} = 1\); 3) \((x + 7)^2 - (x - 2)^2 = 15\); 4) \(5x - 2 = 3(3x - 1) - 4x - 4\); 5) \(6x + (x - 2)(x + 2) = (x + 3)^2 - 13\); 6) \((x + 6)(x - 1) - (x + 3)(x - 4) = 5x\).
1) \(9 - 7(x + 3) = 5 - 6x\)
\(9 - 7x - 21 - 5 + 6x = 0\)
\(- x - 17 = 0\)
\(x = - 17\)
2) \(\frac{x + 3}{2} - \frac{x - 4}{7} = 1\)
\(\dfrac{7(x + 3) - 2(x - 4)}{14} = 1\)
\(7x + 21 - 2x + 8 = 14\)
\(5x = - 15\)
\(x = - 3\)
3) \((x + 7)^{2} - (x - 2)^{2} = 15\)
\(x^{2} + 14x + 49 - x^{2} + 4x - 4 - 15 = 0\)
\(18x + 30 = 0\)
\(x = - \dfrac{30}{18} = - \dfrac{5}{3}\)
4) \(5x - 2 = 3(3x - 1) - 4x - 4\)
\(5x - 2 = 9x - 3 - 4x - 4\)
\(0 = - 5\)
Нет решений.
5) \(6x + (x - 2)(x + 2) = (x + 3)^{2} - 13\)
\(6x + x^{2} - 4 = x^{2} + 6x + 9 - 13\)
\(x^{2} + 6x - 4 = x^{2} + 6x - 4\)
Имеет множество решений: \(x\) – любое число.
6) \((x + 6)(x - 1) - (x + 3)(x - 4) = 5x\)
\(x^{2} + 5x - 6 - \left( x^{2} - x - 12 \right) = 5x\)
\(x^{2} + 5x - 6 - x^{2} + x + 12 = 5x\)
\(x = - 6\)
Ответ: 1) \(x = - 17\); 2) \(x = - 3\); 3) \(x = - \frac{5}{3}\); 4) Нет решений; 5) \(x\) – любое число; 6) \(x = - 6\).
