Страница 196 номер 689, ГДЗ по алгебре за 9 класс к учебнику Мерзляка

1) \(\begin{cases} 2x - y = 13 \\ x^{2} - y^{2} = 23 \end{cases}\ \Rightarrow \begin{cases} y = 2x - 13 \\ x^{2} - (2x - 13)^{2} = 23 \end{cases}\)
\(x^{2} - (2x - 13)^{2} = 23\)
\(x^{2} - 4x^{2} + 52x - 169 = 23\)
\(- 3x^{2} + 52x - 192 = 0\)
\(\dfrac{D}{4} = 676 - 3 \cdot 192 = 676 - 576 = 100\)
\(x_{1} = \dfrac{- 26 - 10}{- 3} = 12;\ y_{1} = 2 \cdot 12 - 13 = 11\)
\(x_{2} = \dfrac{- 26 + 10}{- 3} = \dfrac{16}{3};\ y_{2} = 2 \cdot \dfrac{16}{3} - 13 = \dfrac{32 - 39}{3} = - \dfrac{7}{3}\)

2) \(\begin{cases} 2x^{2} - y^{2} = 23 \\ 2x^{2} + y^{2} = 41 \end{cases}\ \Rightarrow 4x^{2} = 64 \Rightarrow x^{2} = 16 \Rightarrow x = \pm 4\)
\(2x^{2} - y^{2} = 23\)
\(y^{2} = 2x^{2} - 23\)
\(y^{2} = 2 \cdot 16 - 23\)
\(y^{2} = 9\)
\(y = \pm 3\)

Решения \((4;\ 3),( - 4;\ 3),(4; - 3),( - 4; - 3)\)

Ответ: 1) \((12;\ 11),\left( \frac{16}{3}; - \frac{7}{3} \right)\); 2) \((4;\ 3),( - 4;\ 3),(4; - 3),( - 4; - 3)\).