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МатематикаРешебник по алгебре 8 класс Макарычев ФГОС Задание 316
Задание 316
\[\boxed{\text{316\ (}\text{c}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[y = x^{2};\ \ \ - 3 \leq x \leq 3\]
| \[x\] | \[- 3\] | \[- 2\] | \[- 1\] | \[0\] | \[1\] | \[2\] | \[3\] |
|---|---|---|---|---|---|---|---|
| \[y\] | \[9\] | \[4\] | \[1\] | \[0\] | \[1\] | \[4\] | \[9\] |
\[\textbf{а)}\ \text{y\ }( - 2,5) \approx 6,3\]
\[\text{y\ }(1,7) \approx 2,9\]
\[\ б)\ y = 5 \rightarrow x \approx \pm 2,3\]
\[y = 7,5 \rightarrow x \approx \pm 2,8\]
\[\textbf{в)}\ y( - 1,4) \approx 2\]
\[y(2,8) \approx 7,8\]
\[\textbf{г)}\ y = 2,5 \rightarrow x \approx \pm 1,6\]
\[y = 9 \rightarrow x = \pm 3.\]
\[\boxed{\text{316\ (}\text{н}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[Преобразуем\ уравнения\ \]
\[согласно\ определению\ \]
\[арифметического\]
\[квадратного\ корня.\ \]
\[\textbf{а)}\ \sqrt{12 + x} - 7 = 3\]
\[\sqrt{12 + x} = 10\]
\[\left( \sqrt{12 + x} \right)^{2} = 10^{2}\]
\[12 + x = 100\]
\[x = 100 - 12\]
\[x = 88.\]
\[\textbf{б)}\ \sqrt{5x - 1} - 4 = 6\]
\[\sqrt{5x - 1} = 10\]
\[\left( \sqrt{5x - 1} \right)^{2} = 10^{2}\]
\[5x - 1 = 100\]
\[5x = 100 + 1\]
\[5x = 101\]
\[x = \frac{101}{5} = 20,2.\]
\[\textbf{в)}\ 16 - \sqrt{x - 2} = 7\]
\[\sqrt{x - 2} = 16 - 7\]
\[\sqrt{x - 2} = 9\]
\[\left( \sqrt{x - 2} \right)^{2} = 9^{2}\]
\[x - 2 = 81\]
\[x = 83.\]
\[\textbf{г)}\ 12 - \sqrt{3 - 6x} = - 2\]
\[\sqrt{3 - 6x} = 12 + 2\]
\[\sqrt{3 - 6x} = 14\]
\[\left( \sqrt{3 - 6x} \right)^{2} = 14^{2}\]
\[3 - 6x = 196\]
\[6x = 3 - 196\]
\[6x = - 193\]
\[x = - \frac{193}{6}\]
\[x = - 62\frac{1}{6}.\]
\[\boxed{\text{316.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]
Пояснение.
Решение.
\[При\ переносе\ чисел\ из\ одной\ \]
\[части\ уравнения\ в\ другую\ \]
\[меняем\ знак\ \]
\[на\ противоположный.\ \]
\[\textbf{а)}\ 16 + x^{2} = 0\]
\[x^{2} = - 16\]
\[x = \varnothing.\]
\[\textbf{б)}\ 0,3x^{2} = 0,027\]
\[x^{2} = 0,027\ :0,3 = 0,09\]
\[x = \sqrt{0,09} = \pm 0,3.\]
\[\textbf{в)}\ 0,5x^{2} = 30\]
\[x^{2} = 30\ :0,5 = 60\]
\[x = \sqrt{60} = \pm 2\sqrt{15}.\]
\[\textbf{г)} - 5x^{2} = \frac{1}{20}\]
\[x^{2} = \frac{1}{20}\ :( - 5) < 0\]
\[x = \varnothing.\]
\[\textbf{д)}\ x^{3} - 3x = 0\]
\[x\left( x^{2} - 3 \right) = 0\]
\[x = 0\ \ \ \ \ \ \ x^{2} = 3\]
\[x = 0\ \ \ \ \ \ \ x = \pm \sqrt{3}.\]
\[\textbf{е)}\ x^{3} - 11x = 0\]
\[x\left( x^{2} - 11 \right) = 0\]
\[x = 0\ \ \ \ x^{2} = 11\]
\[x = 0\ \ \ \ x = \pm \sqrt{11}.\]
