Страница 79 номер 214, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Решить уравнение (208—223). 1) \(3^{x^2 + x - 12} = 1\); 2) \(2^{x^2 - 7x + 10} = 1\); 3) \(2^{\frac{x - 1}{x - 2}} = 4\); 4) \(0{,}5^{\frac{1}{x}} = 4^{\frac{1}{x + 1}}\).
1) \(3^{x^{2} + x - 12} = 1\)
\(3^{x^{2} + x - 12} = 3^{0},\)
\(x^{2} + x - 12 = 0,\)
\(x_{1{,}2} = \dfrac{-1 \pm \sqrt{1 + 48}}{2},\)
\(x_{1} = 3, x_{2} = -4;\)
2) \(2^{x^{2} - 7x + 10} = 1\)
\(2^{x^{2} - 7x + 10} = 2^{0},\)
\(x^{2} - 7x + 10 = 0,\)
\(x_{1{,}2} = \dfrac{7 \pm \sqrt{49 - 40}}{2},\)
\(x_{1} = 5, x_{2} = 2;\)
3) \(2^{\frac{x - 1}{x - 2}} = 4\)
\(2^{\frac{x - 1}{x - 2}} = 2^{2},\)
\(\dfrac{x - 1}{x - 2} = 2,\)
\(x - 1 = 2x - 4,\)
\(x = 3;\)
4) \(0{,}5^{\frac{1}{x}} = 4^{\frac{1}{x + 1}}\)
\(2^{-\frac{1}{x}} = 2^{\frac{2}{x + 1}},\)
\(-\dfrac{1}{x} = \dfrac{2}{x + 1},\)
\(2x = -x - 1,\)
\(3x = -1,\)
\(x = -\dfrac{1}{3}.\)
