ГДЗ по алгебре 8 класс Макарычев Задание 464

\[\boxed{\text{464\ (464).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ x = 2:\ \ \ \ \ \ \]

\[\sqrt{5x - 10} = \sqrt{5 \cdot 2 - 10} =\]

\[= \sqrt{10 - 10} = \sqrt{0} = 0\]

\[x = 2,2:\ \ \ \]

\[\sqrt{5x - 10} = \sqrt{5 \cdot 2,2 - 10} =\]

\[= \sqrt{11 - 10} = \sqrt{1} = 1\]

\[x = 5,2:\ \ \ \]

\[\sqrt{5x - 10} = \sqrt{5 \cdot 5,2 - 10} =\]

\[= \sqrt{26 - 10} = \sqrt{16} = 4\]

\[x = 22:\ \ \ \ \]

\[\sqrt{5x - 10} = \sqrt{5 \cdot 22 - 10} =\]

\[= \sqrt{110 - 10} = \sqrt{100} = 10\]

\[\textbf{б)}\ y = 1:\ \ \]

\[\sqrt{6 - 2y} = \sqrt{6 - 2 \cdot 1} =\]

\[= \sqrt{6 - 2} = \sqrt{4} = 2\]

\[y = - 1,5:\ \ \]

\[\sqrt{6 - 2y} = \sqrt{6 - 2 \cdot ( - 1,5)} =\]

\[= \sqrt{6 + 3} = \sqrt{9} = 3\]

\[y = - 15:\ \ \]

\[\sqrt{6 - 2y} = \sqrt{6 - 2 \cdot ( - 15)} =\]

\[= \sqrt{6 + 30} = \sqrt{36} = 6\]

\[y = - 37,5:\ \ \]

\[\sqrt{6 - 2y} = \sqrt{6 - 2 \cdot ( - 37,5)} =\]

\[= \sqrt{6 + 75} = \sqrt{81} = 9\]

\[\textbf{в)}\ x = 0:\ \ \ \]

\[\frac{3 + \sqrt{x}}{3 - \sqrt{x}} = \frac{3 + \sqrt{0}}{3 - \sqrt{0}} = \frac{3}{3} = 1\]

\[x = 1:\ \ \ \]

\[\frac{3 + \sqrt{x}}{3 - \sqrt{x}} = \frac{3 + \sqrt{1}}{3 - \sqrt{1}} = \frac{4}{2} = 2\]

\[x = 16:\ \ \ \]

\[\frac{3 + \sqrt{x}}{3 - \sqrt{x}} = \frac{3 + \sqrt{16}}{3 - \sqrt{16}} = \frac{3 + 4}{3 - 4} =\]

\[= - \frac{7}{1} = - 7\]

\[x = 0,25:\ \ \ \]

\[\frac{3 + \sqrt{x}}{3 - \sqrt{x}} = \frac{3 + \sqrt{0,25}}{3 - \sqrt{0,25}} = \frac{3 + 0,5}{3 - 0,5} =\]

\[= \frac{3,5}{2,5} = \frac{7}{5} = 1,4\]

\[\textbf{г)}\ a = 0,\ \ \ b = 0:\ \ \ \]

\[\sqrt{2a - b} = \sqrt{2 \cdot 0 - 0} = \sqrt{0} = 0\]

\[a = 4,\ \ \ b = 7:\ \ \ \ \ \ \ \]

\[\ \sqrt{2a - b} = \sqrt{2 \cdot 4 - 7} =\]

\[= \sqrt{8 - 7} = \sqrt{1} = 1\ \ \ \ \ \]