Решебник по алгебре 8 класс Макарычев ФГОС Задание 316

\[\boxed{\text{316\ (}\text{c}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[y = x^{2};\ \ \ - 3 \leq x \leq 3\]

\[\textbf{а)}\ \text{y\ }( - 2,5) \approx 6,3\]

\[\text{y\ }(1,7) \approx 2,9\]

\[\ б)\ y = 5 \rightarrow x \approx \pm 2,3\]

\[y = 7,5 \rightarrow x \approx \pm 2,8\]

\[\textbf{в)}\ y( - 1,4) \approx 2\]

\[y(2,8) \approx 7,8\]

\[\textbf{г)}\ y = 2,5 \rightarrow x \approx \pm 1,6\]

\[y = 9 \rightarrow x = \pm 3.\]

\[\boxed{\text{316\ (}\text{н}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[Преобразуем\ уравнения\ \]

\[согласно\ определению\ \]

\[арифметического\]

\[квадратного\ корня.\ \]

\[\textbf{а)}\ \sqrt{12 + x} - 7 = 3\]

\[\sqrt{12 + x} = 10\]

\[\left( \sqrt{12 + x} \right)^{2} = 10^{2}\]

\[12 + x = 100\]

\[x = 100 - 12\]

\[x = 88.\]

\[\textbf{б)}\ \sqrt{5x - 1} - 4 = 6\]

\[\sqrt{5x - 1} = 10\]

\[\left( \sqrt{5x - 1} \right)^{2} = 10^{2}\]

\[5x - 1 = 100\]

\[5x = 100 + 1\]

\[5x = 101\]

\[x = \frac{101}{5} = 20,2.\]

\[\textbf{в)}\ 16 - \sqrt{x - 2} = 7\]

\[\sqrt{x - 2} = 16 - 7\]

\[\sqrt{x - 2} = 9\]

\[\left( \sqrt{x - 2} \right)^{2} = 9^{2}\]

\[x - 2 = 81\]

\[x = 83.\]

\[\textbf{г)}\ 12 - \sqrt{3 - 6x} = - 2\]

\[\sqrt{3 - 6x} = 12 + 2\]

\[\sqrt{3 - 6x} = 14\]

\[\left( \sqrt{3 - 6x} \right)^{2} = 14^{2}\]

\[3 - 6x = 196\]

\[6x = 3 - 196\]

\[6x = - 193\]

\[x = - \frac{193}{6}\]

\[x = - 62\frac{1}{6}.\]

\[\boxed{\text{316.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]

\[При\ переносе\ чисел\ из\ одной\ \]

\[части\ уравнения\ в\ другую\ \]

\[меняем\ знак\ \]

\[на\ противоположный.\ \]

\[\textbf{а)}\ 16 + x^{2} = 0\]

\[x^{2} = - 16\]

\[x = \varnothing.\]

\[\textbf{б)}\ 0,3x^{2} = 0,027\]

\[x^{2} = 0,027\ :0,3 = 0,09\]

\[x = \sqrt{0,09} = \pm 0,3.\]

\[\textbf{в)}\ 0,5x^{2} = 30\]

\[x^{2} = 30\ :0,5 = 60\]

\[x = \sqrt{60} = \pm 2\sqrt{15}.\]

\[\textbf{г)} - 5x^{2} = \frac{1}{20}\]

\[x^{2} = \frac{1}{20}\ :( - 5) < 0\]

\[x = \varnothing.\]

\[\textbf{д)}\ x^{3} - 3x = 0\]

\[x\left( x^{2} - 3 \right) = 0\]

\[x = 0\ \ \ \ \ \ \ x^{2} = 3\]

\[x = 0\ \ \ \ \ \ \ x = \pm \sqrt{3}.\]

\[\textbf{е)}\ x^{3} - 11x = 0\]

\[x\left( x^{2} - 11 \right) = 0\]

\[x = 0\ \ \ \ x^{2} = 11\]

\[x = 0\ \ \ \ x = \pm \sqrt{11}.\]