Страница 65 номер 2.15, ГДЗ по алгебре за 8 класс к учебнику Дорофеева
Найдите значение выражения при \(a = 9\); 1; \(\frac{1}{4}\):
а) \(a + \sqrt{a}\); б) \(a - \sqrt{a}\); в) \(a\sqrt{a}\); г) \(\frac{a}{\sqrt{a}}\).
Так как \(3^2 = 9\), \(1^2 = 1\) и \(\left(\frac{1}{2}\right)^2 = \frac{1}{4}\), то \(\sqrt{9} = 3\), \(\sqrt{1} = 1\) и \(\sqrt{\frac{1}{4}} = \frac{1}{2}\).
а) \[\begin{aligned} 9 + \sqrt{9} &= 9 + 3 = 12; \\ 1 + \sqrt{1} &= 1 + 1 = 2; \\ \frac{1}{4} + \sqrt{\frac{1}{4}} &= \frac{1}{4} + \frac{1}{2} = \frac{1}{4} + \frac{2}{4} = \frac{3}{4}. \end{aligned}\]
б) \[\begin{aligned} 9 - \sqrt{9} &= 9 - 3 = 6; \\ 1 - \sqrt{1} &= 1 - 1 = 0; \\ \frac{1}{4} - \sqrt{\frac{1}{4}} &= \frac{1}{4} - \frac{1}{2} = \frac{1}{4} - \frac{2}{4} = -\frac{1}{4}. \end{aligned}\]
в) \[\begin{aligned} 9\sqrt{9} &= 9 \cdot 3 = 27; \\ 1 \cdot \sqrt{1} &= 1 \cdot 1 = 1; \\ \frac{1}{4}\sqrt{\frac{1}{4}} &= \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8}. \end{aligned}\]
г) \[\begin{aligned} \frac{9}{\sqrt{9}} &= \frac{9}{3} = 3; \\ \frac{1}{\sqrt{1}} &= \frac{1}{1} = 1; \\ \frac{\frac{1}{4}}{\sqrt{\frac{1}{4}}} &= \frac{1}{4} : \frac{1}{2} = \frac{1}{4} \cdot 2 = \frac{1}{2}. \end{aligned}\]
Ответ: при \(a = 9\); 1; \(\frac{1}{4}\) соответственно: а) 12; 2; \(\frac{3}{4}\); б) 6; 0; \(-\frac{1}{4}\); в) 27; 1; \(\frac{1}{8}\); г) 3; 1; \(\frac{1}{2}\).
Найдите значение выражения при \(a = 9\); 1; \(\frac{1}{4}\):
а) \(a + \sqrt{a}\); б) \(a - \sqrt{a}\); в) \(a\sqrt{a}\); г) \(\frac{a}{\sqrt{a}}\).
Так как \(3^2 = 9\), \(1^2 = 1\) и \(\left(\frac{1}{2}\right)^2 = \frac{1}{4}\), то \(\sqrt{9} = 3\), \(\sqrt{1} = 1\) и \(\sqrt{\frac{1}{4}} = \frac{1}{2}\). Подставим в каждое выражение \(a = 9\), \(a = 1\) и \(a = \frac{1}{4}\).
а) \[\begin{aligned} 9 + \sqrt{9} &= 9 + 3 = 12; \\ 1 + \sqrt{1} &= 1 + 1 = 2; \\ \frac{1}{4} + \sqrt{\frac{1}{4}} &= \frac{1}{4} + \frac{1}{2} = \frac{1}{4} + \frac{2}{4} = \frac{3}{4}. \end{aligned}\]
б) \[\begin{aligned} 9 - \sqrt{9} &= 9 - 3 = 6; \\ 1 - \sqrt{1} &= 1 - 1 = 0; \\ \frac{1}{4} - \sqrt{\frac{1}{4}} &= \frac{1}{4} - \frac{1}{2} = \frac{1}{4} - \frac{2}{4} = -\frac{1}{4}. \end{aligned}\]
в) \[\begin{aligned} 9\sqrt{9} &= 9 \cdot 3 = 27; \\ 1 \cdot \sqrt{1} &= 1 \cdot 1 = 1; \\ \frac{1}{4}\sqrt{\frac{1}{4}} &= \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8}. \end{aligned}\]
г) \[\begin{aligned} \frac{9}{\sqrt{9}} &= \frac{9}{3} = 3; \\ \frac{1}{\sqrt{1}} &= \frac{1}{1} = 1; \\ \frac{\frac{1}{4}}{\sqrt{\frac{1}{4}}} &= \frac{1}{4} : \frac{1}{2} = \frac{1}{4} \cdot 2 = \frac{1}{2}. \end{aligned}\]
Ответ: при \(a = 9\); 1; \(\frac{1}{4}\) соответственно: а) 12; 2; \(\frac{3}{4}\); б) 6; 0; \(-\frac{1}{4}\); в) 27; 1; \(\frac{1}{8}\); г) 3; 1; \(\frac{1}{2}\).
