Страница 98 номер 2.136, ГДЗ по алгебре за 8 класс к учебнику Дорофеева
Найдите значения выражений \(\frac{xy}{x+y}\) и \(\frac{x-y}{xy}\) при:
а) \(x=\sqrt{2}\), \(y=\sqrt{8}\);
б) \(x=2-\sqrt{3}\), \(y=2+\sqrt{3}\);
в) \(x=\sqrt{6}-\sqrt{3}\), \(y=\sqrt{6}+\sqrt{3}\);
г) \(x=\sqrt{5}+\sqrt{2}\), \(y=\sqrt{5}-\sqrt{2}\).
а) \(\sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2}\). \[\begin{aligned} xy &= \sqrt{2} \cdot \sqrt{8} = \sqrt{16} = 4; \\ x + y &= \sqrt{2} + 2\sqrt{2} = 3\sqrt{2}; \\ x - y &= \sqrt{2} - 2\sqrt{2} = -\sqrt{2}. \end{aligned}\]
Значит, \[\begin{aligned} \frac{xy}{x + y} &= \frac{4}{3\sqrt{2}} = \frac{4\sqrt{2}}{3(\sqrt{2})^2} = \frac{4\sqrt{2}}{6} = \frac{2\sqrt{2}}{3}; \\ \frac{x - y}{xy} &= \frac{-\sqrt{2}}{4} = -\frac{\sqrt{2}}{4}. \end{aligned}\]
б) \[\begin{aligned} xy &= (2 - \sqrt{3})(2 + \sqrt{3}) = 4 - 3 = 1; \\ x + y &= (2 - \sqrt{3}) + (2 + \sqrt{3}) = 4; \\ x - y &= (2 - \sqrt{3}) - (2 + \sqrt{3}) = -2\sqrt{3}. \end{aligned}\]
Значит, \[\begin{aligned} \frac{xy}{x + y} &= \frac{1}{4}; \\ \frac{x - y}{xy} &= \frac{-2\sqrt{3}}{1} = -2\sqrt{3}. \end{aligned}\]
в) \[\begin{aligned} xy &= (\sqrt{6} - \sqrt{3})(\sqrt{6} + \sqrt{3}) = 6 - 3 = 3; \\ x + y &= (\sqrt{6} - \sqrt{3}) + (\sqrt{6} + \sqrt{3}) = 2\sqrt{6}; \\ x - y &= (\sqrt{6} - \sqrt{3}) - (\sqrt{6} + \sqrt{3}) = -2\sqrt{3}. \end{aligned}\]
Значит, \[\begin{aligned} \frac{xy}{x + y} &= \frac{3}{2\sqrt{6}}; \\ \frac{x - y}{xy} &= \frac{-2\sqrt{3}}{3} = -\frac{2\sqrt{3}}{3}. \end{aligned}\]
г) \[\begin{aligned} xy &= (\sqrt{5} + \sqrt{2})(\sqrt{5} - \sqrt{2}) = 5 - 2 = 3; \\ x + y &= (\sqrt{5} + \sqrt{2}) + (\sqrt{5} - \sqrt{2}) = 2\sqrt{5}; \\ x - y &= (\sqrt{5} + \sqrt{2}) - (\sqrt{5} - \sqrt{2}) = 2\sqrt{2}. \end{aligned}\]
Значит, \[\begin{aligned} \frac{xy}{x + y} &= \frac{3}{2\sqrt{5}}; \\ \frac{x - y}{xy} &= \frac{2\sqrt{2}}{3}. \end{aligned}\]
Ответ: а) \(\frac{2\sqrt{2}}{3}\) и \(-\frac{\sqrt{2}}{4}\); б) \(\frac{1}{4}\) и \(-2\sqrt{3}\); в) \(\frac{3}{2\sqrt{6}}\) и \(-\frac{2\sqrt{3}}{3}\); г) \(\frac{3}{2\sqrt{5}}\) и \(\frac{2\sqrt{2}}{3}\).
Найдите значения выражений \(\frac{xy}{x+y}\) и \(\frac{x-y}{xy}\) при:
а) \(x=\sqrt{2}\), \(y=\sqrt{8}\);
б) \(x=2-\sqrt{3}\), \(y=2+\sqrt{3}\);
в) \(x=\sqrt{6}-\sqrt{3}\), \(y=\sqrt{6}+\sqrt{3}\);
г) \(x=\sqrt{5}+\sqrt{2}\), \(y=\sqrt{5}-\sqrt{2}\).
Обе дроби составлены из произведения \(xy\), суммы \(x + y\) и разности \(x - y\), поэтому найдём их заранее.
а) При \(x = \sqrt{2}\), \(y = \sqrt{8}\) имеем \(\sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2}\) и \[\begin{aligned} xy &= \sqrt{2} \cdot \sqrt{8} = \sqrt{16} = 4; \\ x + y &= \sqrt{2} + 2\sqrt{2} = 3\sqrt{2}; \\ x - y &= \sqrt{2} - 2\sqrt{2} = -\sqrt{2}. \end{aligned}\]
В первой дроби освободимся от иррациональности в знаменателе — умножим её числитель и знаменатель на \(\sqrt{2}\): \[\begin{aligned} \frac{xy}{x + y} &= \frac{4}{3\sqrt{2}} = \frac{4\sqrt{2}}{3(\sqrt{2})^2} = \frac{4\sqrt{2}}{6} = \frac{2\sqrt{2}}{3}; \\ \frac{x - y}{xy} &= \frac{-\sqrt{2}}{4} = -\frac{\sqrt{2}}{4}. \end{aligned}\]
б) При \(x = 2 - \sqrt{3}\), \(y = 2 + \sqrt{3}\): \[\begin{aligned} xy &= (2 - \sqrt{3})(2 + \sqrt{3}) = 4 - 3 = 1; \\ x + y &= (2 - \sqrt{3}) + (2 + \sqrt{3}) = 4; \\ x - y &= (2 - \sqrt{3}) - (2 + \sqrt{3}) = -2\sqrt{3}. \end{aligned}\]
Значит, \[\begin{aligned} \frac{xy}{x + y} &= \frac{1}{4}; \\ \frac{x - y}{xy} &= \frac{-2\sqrt{3}}{1} = -2\sqrt{3}. \end{aligned}\]
в) При \(x = \sqrt{6} - \sqrt{3}\), \(y = \sqrt{6} + \sqrt{3}\): \[\begin{aligned} xy &= (\sqrt{6} - \sqrt{3})(\sqrt{6} + \sqrt{3}) = 6 - 3 = 3; \\ x + y &= (\sqrt{6} - \sqrt{3}) + (\sqrt{6} + \sqrt{3}) = 2\sqrt{6}; \\ x - y &= (\sqrt{6} - \sqrt{3}) - (\sqrt{6} + \sqrt{3}) = -2\sqrt{3}. \end{aligned}\]
Значит, \[\begin{aligned} \frac{xy}{x + y} &= \frac{3}{2\sqrt{6}}; \\ \frac{x - y}{xy} &= \frac{-2\sqrt{3}}{3} = -\frac{2\sqrt{3}}{3}. \end{aligned}\]
г) При \(x = \sqrt{5} + \sqrt{2}\), \(y = \sqrt{5} - \sqrt{2}\): \[\begin{aligned} xy &= (\sqrt{5} + \sqrt{2})(\sqrt{5} - \sqrt{2}) = 5 - 2 = 3; \\ x + y &= (\sqrt{5} + \sqrt{2}) + (\sqrt{5} - \sqrt{2}) = 2\sqrt{5}; \\ x - y &= (\sqrt{5} + \sqrt{2}) - (\sqrt{5} - \sqrt{2}) = 2\sqrt{2}. \end{aligned}\]
Значит, \[\begin{aligned} \frac{xy}{x + y} &= \frac{3}{2\sqrt{5}}; \\ \frac{x - y}{xy} &= \frac{2\sqrt{2}}{3}. \end{aligned}\]
Ответ: а) \(\frac{2\sqrt{2}}{3}\) и \(-\frac{\sqrt{2}}{4}\); б) \(\frac{1}{4}\) и \(-2\sqrt{3}\); в) \(\frac{3}{2\sqrt{6}}\) и \(-\frac{2\sqrt{3}}{3}\); г) \(\frac{3}{2\sqrt{5}}\) и \(\frac{2\sqrt{2}}{3}\).
