Решебник по алгебре 9 класс Мерзляк Задание 988

\[\boxed{\mathbf{988\ (988).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[По\ т.\ Пифагора\ a^{2} + b^{2} = d^{2}\text{\ \ }\]

\[или\ \ \ a^{2} + b^{2} = 17^{2}.\]

\[S = a \cdot b = 120.\]

\[Составим\ систему:\]

\[\left\{ \begin{matrix} ab = 120\ \ \ | \cdot 2 \\ a^{2} + b^{2} = 289\ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 2ab = 240\ \ \ \ \ \ \ \\ a^{2} + b^{2} = 289 \\ \end{matrix} \right.\ \ +\]

\[a^{2} + 2ab + b^{2} = 529\]

\[(a + b)^{2} = 529\]

\[a + b = 23\ \ \ \ или\ \ \ \ \ \]

\[a + b =\]

\[= - 23\ (не\ удовлетворяет).\]

\[\left\{ \begin{matrix} a + b = 23 \\ ab = 120\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} a = 23 - b\ \ \ \ \ \ \ \ \ \ \\ (23 - b)b = 120 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\text{\ \ }\left\{ \begin{matrix} a = 23 - b\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ b^{2} - 23b + 120 = 0 \\ \end{matrix} \right.\ \]

\[b_{1} + b_{2} = 23\]

\[b_{1} \cdot b_{2} = 120\]

\[\left\{ \begin{matrix} a = 23 - b \\ b = 15\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} a = 8 \\ b = 15 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\text{\ \ }или\ \ \ \ \ \ \left\{ \begin{matrix} a = 15 \\ b = 8\ \ \\ \end{matrix} \right.\ \]

\[Ответ:8\ см,\ 15\ см.\]