Страница 256 номер 864, ГДЗ по алгебре за 10-11 класс Алимов - ответы и решение

1) \(y = 8 - x\) и \(y = 4\sqrt{x + 4}\)
\(8 - x = 4\sqrt{x + 4}, 64 - 16x + x^{2} = 16x + 64,\)
\(x^{2} - 32x = 0, x_{1} = 0,\)
\(x_{2} = 32\) – не подходит, т.к. \(8 - x > 0\).
\(\operatorname{tg} \alpha_{1} = y'_{1}(0) = -1, \alpha_{1} = \dfrac{3\pi}{4};\)
\(\operatorname{tg} \alpha_{2} = y'_{2}(0) = \dfrac{2}{\sqrt{0 + 4}} = 1, \alpha_{2} = \dfrac{\pi}{4}\)
\(\beta = \dfrac{3\pi}{4} - \dfrac{\pi}{4} = \dfrac{\pi}{2};\)

2) \(y = \frac{1}{2}(x + 1)^{2}\) и \(y = \frac{1}{2}(x - 1)^{2}\)
\(\dfrac{1}{2}(x + 1)^{2} = \dfrac{1}{2}(x - 1)^{2}\)
\(x^{2} + 2x + 1 = x^{2} - 2x + 1, 4x = 0, x = 0\)
\(\operatorname{tg} \alpha_{1} = y'_{1}(0) = 0 + 1 = 1, \alpha_{1} = \dfrac{\pi}{4};\)
\(\operatorname{tg} \alpha_{2} = y'_{2}(0) = 0 - 1 = -1, \alpha_{2} = \dfrac{3\pi}{4}\)
\(\beta = \dfrac{3\pi}{4} - \dfrac{\pi}{4} = \dfrac{\pi}{2};\)

3) \(y = \ln(1 + x)\) и \(y = \ln(1 - x)\)
\(\ln(1 + x) = \ln(1 - x), 1 + x = 1 - x, 2x = 0, x = 0\)
\(\operatorname{tg} \alpha_{1} = y'_{1}(0) = \dfrac{1}{1 + 0} = 1, \alpha_{1} = \dfrac{\pi}{4};\)
\(\operatorname{tg} \alpha_{2} = y'_{2}(0) = \dfrac{-1}{1 - 0} = -1, \alpha_{2} = \dfrac{3\pi}{4}\)
\(\beta = \dfrac{3\pi}{4} - \dfrac{\pi}{4} = \dfrac{\pi}{2};\)

4) \(y = e^{x}\) и \(y = e^{-x}\)
\(e^{x} = e^{-x}\)
\(x = -x\)
\(x = 0\)
\(\operatorname{tg} \alpha_{1} = y'_{1}(0) = e^{0} = 1, \alpha_{1} = \dfrac{\pi}{4};\)
\(\operatorname{tg} \alpha_{2} = y'_{2}(0) = -e^{0} = -1, \alpha_{2} = \dfrac{3\pi}{4}\)
\(\beta = \dfrac{3\pi}{4} - \dfrac{\pi}{4} = \dfrac{\pi}{2}.\)