Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 222

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\[1)\ \left\lbrack \begin{matrix} x^{2} + 4x + 4 = 0 \\ 2x^{2} - x - 1 = 0 \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[x^{2} + 4x + 4 = 0\]

\[(x + 2)^{2} = 0\]

\[x + 2 = 0\]

\[x = - 2.\]

\[2x^{2} - x - 1 = 0\]

\[D = 1 + 8 = 9\]

\[x_{1} = \frac{1 + 3}{4} = 1;\ \ \]

\[x_{2} = \frac{1 - 3}{4} = - 0,5.\]

\[Ответ:x = \left\{ - 2;\ - 0,5;1 \right\}.\]

\[2)\ \left\lbrack \begin{matrix} |x - 3| = 2\ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 7x + 10 = 0 \\ \end{matrix} \right.\ \]

\[|x - 3| = 2\]

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

\[x = 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 1.\]

\[x^{2} - 7x + 10 = 0\]

\[x_{1} + x_{2} = 7;\ \ \ x_{1} \cdot x_{2} = 10\]

\[x_{1} = 5;\ \ \ x_{2} = 2.\]

\[Ответ:x = \left\{ 1;2;5 \right\}.\]