ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 315

 

\[\boxed{\text{315\ (}\text{c}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\(\left( \sqrt{n^{2} + 39} \right)^{2} = x^{2}\)

\[n^{2} + 39 = x^{2}\]

\[x^{2} - n^{2} = 39\]

\[(x - n) \cdot (x + n) = 39\]

\[x - двузначное\ число;\ \ x \geq 0;\ \ \]

\[n \in N.\]

\[Ответ:при\ n = 19.\]

\[\boxed{\text{315\ (}\text{н}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[Преобразуем\ уравнения\ \]

\[согласно\ определению\ \]

\[арифметического\]

\[квадратного\ корня.\ \]

\[\textbf{а)}\ \sqrt{3x - 1} = 1\]

\[\left( \sqrt{3x - 1} \right)^{2} = 1^{2}\]

\[3x - 1 = 1\]

\[3x = 1 + 1\]

\[3x = 2\]

\[x = \frac{2}{3}.\]

\[\textbf{б)}\ \sqrt{6x + 4} = 2\]

\[\left( \sqrt{6x + 4} \right)^{2} = 2^{2}\]

\[6x + 4 = 4\]

\[6x = 0\]

\[x = 0.\]

\[\textbf{в)}\ \sqrt{12 - x} = 6\]

\[\left( \sqrt{12 - x} \right)^{2} = 6^{2}\]

\[12 - x = 36\]

\[x = 12 - 36\]

\[x = - 24.\]

\[\textbf{г)}\ \sqrt{8x - 1} = 1\]

\[\left( \sqrt{8x - 1} \right)^{2} = 1^{2}\]

\[8x - 1 = 1\]

\[8x = 2\]

\[x = \frac{2}{8} = \frac{1}{4}\]

\[x = 0,25.\]

\[\boxed{\text{315.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

Пояснение.

Решение.

\[Перенесем\ числа\ в\ левую\ часть\ \]

\[уравнения,\ меняя\ знак\ на\ \]

\[противоположный.\]

\[При\ переносе\ буквенной\ части,\ \]

\[тоже\ меняем\ знак\ на\ \]

\[противоположный.\]

\[\textbf{а)}\ 80 + y^{2} = 81\]

\[y^{2} = 81 - 80\]

\[y^{2} = 1\]

\[y = \sqrt{1}\]

\[y = \pm 1.\]

\[\textbf{б)}\ 19 + c^{2} = 10\]

\[c^{2} = 10 - 19\]

\[c = - 9 < 0\]

\(c = \varnothing\).

\[\textbf{в)}\ 20 - b^{2} = - 5\]

\[b^{2} = 20 + 5\]

\[b^{2} = 25\]

\[b = \sqrt{25}\]

\[b = \pm 5.\]

\[\textbf{г)}\ 3x^{2} = 1,47\]

\[x^{2} = 1,47\ :3\]

\[x^{2} = 0,9\]

\[x = \sqrt{0,49}\]

\[x = \pm 0,7.\]

\[\textbf{д)}\frac{1}{4}a^{2} = 10\]

\[a^{2} = 40\]

\[a = \sqrt{40}\]

\[a = \pm 2\sqrt{10}.\]

\[\textbf{е)} - 5y^{2} = 1,8\]

\[y^{2} = 1,8\ :( - 5) < 0\]

\[y = \varnothing.\]