\[\boxed{\text{316\ (}\text{c}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[y = x^{2};\ \ \ - 3 \leq x \leq 3\]
\[x\] | \[- 3\] | \[- 2\] | \[- 1\] | \[0\] | \[1\] | \[2\] | \[3\] |
---|---|---|---|---|---|---|---|
\[y\] | \[9\] | \[4\] | \[1\] | \[0\] | \[1\] | \[4\] | \[9\] |
\[\textbf{а)}\ \text{y\ }( - 2,5) \approx 6,3\]
\[\text{y\ }(1,7) \approx 2,9\]
\[\ б)\ y = 5 \rightarrow x \approx \pm 2,3\]
\[y = 7,5 \rightarrow x \approx \pm 2,8\]
\[\textbf{в)}\ y( - 1,4) \approx 2\]
\[y(2,8) \approx 7,8\]
\[\textbf{г)}\ y = 2,5 \rightarrow x \approx \pm 1,6\]
\[y = 9 \rightarrow x = \pm 3.\]
\[\boxed{\text{316\ (}\text{н}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[Преобразуем\ уравнения\ \]
\[согласно\ определению\ \]
\[арифметического\]
\[квадратного\ корня.\ \]
\[\textbf{а)}\ \sqrt{12 + x} - 7 = 3\]
\[\sqrt{12 + x} = 10\]
\[\left( \sqrt{12 + x} \right)^{2} = 10^{2}\]
\[12 + x = 100\]
\[x = 100 - 12\]
\[x = 88.\]
\[\textbf{б)}\ \sqrt{5x - 1} - 4 = 6\]
\[\sqrt{5x - 1} = 10\]
\[\left( \sqrt{5x - 1} \right)^{2} = 10^{2}\]
\[5x - 1 = 100\]
\[5x = 100 + 1\]
\[5x = 101\]
\[x = \frac{101}{5} = 20,2.\]
\[\textbf{в)}\ 16 - \sqrt{x - 2} = 7\]
\[\sqrt{x - 2} = 16 - 7\]
\[\sqrt{x - 2} = 9\]
\[\left( \sqrt{x - 2} \right)^{2} = 9^{2}\]
\[x - 2 = 81\]
\[x = 83.\]
\[\textbf{г)}\ 12 - \sqrt{3 - 6x} = - 2\]
\[\sqrt{3 - 6x} = 12 + 2\]
\[\sqrt{3 - 6x} = 14\]
\[\left( \sqrt{3 - 6x} \right)^{2} = 14^{2}\]
\[3 - 6x = 196\]
\[6x = 3 - 196\]
\[6x = - 193\]
\[x = - \frac{193}{6}\]
\[x = - 62\frac{1}{6}.\]