Решите уравнение: x^2+18x-63=0

\[\ x^{2} + 18x - 63 = 0\]

\[x_{1} + x_{2} = - 18\]

\[x_{1} \cdot x_{2} = - 63 \Longrightarrow x_{1} = - 21\ \ и\ \ \ \ x_{2} = 3\]

\[Ответ:\ \ x_{1} = - 21\ \ и\ \ \ \ x_{2} = 3.\]

\[Пусть\ b\ см - одна\ сторона\ \]

\[прямоугольника.\]

\[По\ условию\ задачи,\ периметр\ 22\ см\ и\ \]

\[площадь\ равна\ 24\ см^{2}.\]

\[Составим\ уравнение:\]

\[2 \cdot \left( b + \frac{24}{b} \right) = 22\]

\[b^{\backslash b} + \frac{24}{b} = 11^{\backslash b}\]

\[b^{2} + 24 - 11b = 0\]

\[b^{2} - 11b + 24 = 0\]

\[b_{1} + b_{2} = 11\]

\[b_{1} \cdot b_{2} = 24 \Longrightarrow b_{1} = 8\ \ и\ \ \ b_{2} = 3.\]

\[3)\ 11 - 8 = 3\ см.\]

\[4)\ 11 - 3 = 8\ см.\]

\[Ответ:стороны\ равны\ 3\ см\ и\ 8\ см.\]

\[x^{2} - 7x + q = 0\ \ \ и\ \ \ x_{1} = 13\]

\[x_{1} + x_{2} = 7\]

\[13 + x_{2} = 7\]

\[x_{2} = - 6.\]

\[x_{1} \cdot x_{2} = q\]

\[- 6 \cdot 13 = q\]

\[q = - 78.\]

\[Ответ:\ \ x_{2} = - 6\ \ и\ \ \ q = - 78.\]

\[\ \frac{x^{2}}{x^{2} - 9} = \frac{12 - x}{x^{2} - 9}\]

\[\frac{x^{2}}{x^{2} - 9} - \frac{12 - x}{x^{2} - 9} = 0\]

\[\frac{x^{2} + x - 12}{x^{2} - 9} = 0;\ \ \ \ \ \ \ x \neq \pm 3\]

\[x^{2} + x - 12 = 0\]

\[x_{1} + x_{2} = - 1;\ \ \ x_{1} \cdot x_{2} = - 12\]

\[x_{1} = - 4;\ \ \ x_{2} = 3\ (не\ подходит).\]

\[Ответ:x = - 4.\]