Страница 88 номер 250, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение
Решить уравнение (250—252). 1) \(1{,}5^{5x - 7} = \left(\frac{2}{3}\right)^{x + 1}\); 2) \(0{,}75^{2x - 3} = \left(1\frac{1}{3}\right)^{5 - x}\); 3) \(5^{x^2 - 5x - 6} = 1\); 4) \(\left(\frac{1}{7}\right)^{x^2 - 2x - 2} = \frac{1}{7}\).
1) \(1{,}5^{5x-7} = \left(\frac{2}{3}\right)^{x+1}\),
\(\left(\dfrac{3}{2}\right)^{5x-7} = \left(\dfrac{2}{3}\right)^{x+1},\)
\(7 - 5x = x + 1,\)
\(6x = 6,\)
\(x = 1;\)
2) \(0{,}75^{2x-3} = \left(1\frac{1}{3}\right)^{5-x}\),
\(\left(\dfrac{3}{4}\right)^{2x-3} = \left(\dfrac{4}{3}\right)^{5-x},\)
\(2x - 3 = x - 5,\)
\(x = -2;\)
3) \(5^{x^{2}-5x-6} = 1\),
\(x^{2} - 5x - 6 = 0,\)
\(x_{1{,}2} = \dfrac{5 \pm \sqrt{25+24}}{2},\)
\(x_{1} = 6, x_{2} = -1;\)
4) \(\left(\frac{1}{7}\right)^{x^{2}-2x-2} = \frac{1}{7}\),
\(x^{2} - 2x - 3 = 0,\)
\(x_{1{,}2} = \dfrac{2 \pm \sqrt{4+12}}{2},\)
\(x_{1} = 3, x_{2} = -1.\)
