Решебник по алгебре 8 класс Мерзляк ФГОС Задание 544

 

\[\boxed{\mathbf{544\ (544).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ a² - 3 = (a - \sqrt{3})(a + \sqrt{3})\]

\[2)\ 4b² - 2 =\]

\[= (2b - \sqrt{2})(2b + \sqrt{2})\]

\[3)\ 5 - 6c^{2} =\]

\[= (\sqrt{5} - c\sqrt{6})(\sqrt{5} + c\sqrt{6})\]

\[4)\ a - 9 = (\sqrt{a} - 3)(\sqrt{a} + 3)\]

\[5)\ m - n =\]

\[= (\sqrt{m} - \sqrt{n})(\sqrt{m} + \sqrt{n})\]

\[6)\ 16x - 25y =\]

\[= (4\sqrt{x} - 5\sqrt{y})(4\sqrt{x} + 5\sqrt{y})\]

\[7)\ a - 2\sqrt{a} + 1 = (\sqrt{a} - 1)²\]

\[8)\ 4m - 28\sqrt{\text{mn}} + 49n =\]

\[= (2\sqrt{m} - 7\sqrt{n})²\]

\[9)\ b + 6\sqrt{b} + 9 = (\sqrt{b} + 3)²\]

\[10)\ 3 + 2\sqrt{3c} + c = (\sqrt{3} + \sqrt{c})²\]

\[11)\ 2 + \sqrt{2} = \sqrt{2} \cdot \sqrt{2} + \sqrt{2} =\]

\[= \sqrt{2} \cdot (\sqrt{2} + 1)\]

\[12)\ 6\sqrt{7} - 7 = 6\sqrt{7} - \sqrt{7} \cdot \sqrt{7} =\]

\[= \sqrt{7} \cdot (6 - \sqrt{7})\]

\[13)\ a - \sqrt{a} = \sqrt{a} \cdot (\sqrt{a} - 1)\]

\[14)\ \sqrt{b} + \sqrt{3b} = \sqrt{b} \cdot (1 + \sqrt{3})\]

\[15)\ \sqrt{15} - \sqrt{5} = \sqrt{5} \cdot (\sqrt{3} - 1)\]

\[\boxed{\mathbf{5}\mathbf{4}\mathbf{4}\mathbf{\text{.\ }}Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ \sqrt{x^{2}} = x - 4\]

\[|x| = x - 4\]

\[x - x = - 4,\text{\ \ \ \ \ } - x - x = - 4\]

\[x = 2,\ \ но\ 2 \neq - 2.\]

\[Ответ:нет\ таких\text{\ \ x}.\]

\[2)\ \sqrt{x^{2}} = 6 - x\]

\[|x| = 6 - x\]

\[- x = 6 - x,\ \ 0 \neq 6\]

\[x = 6 - x,\ \ x = 3\]

\[Ответ:x = 3.\]

\[3)\ 2\sqrt{x^{2}} = x + 3\]

\[2|x| = x + 3\]

\[2x - x = 3,\ \ - 2x - x = 3\]

\[x = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = - 1\]

\[Ответ:\ \ x = 3;\ x = - 1.\]