Страница 129 номер 468, ГДЗ по алгебре за 9 класс к учебнику Мерзляка

1) \(\begin{cases} x^{2} - 12xy + 36y^{2} = 36 \\ x + 6y = 8 \end{cases}\ \Rightarrow \begin{cases} (x - 6y)^{2} = 36 \\ x + 6y = 8 \end{cases}\ \Rightarrow \begin{cases} x - 6y = \pm 6 \\ x + 6y = 8 \end{cases}\)
\(\begin{cases} x - 6y = 6 \\ x + 6y = 8 \end{cases}\ \Rightarrow 2x = 14 \Rightarrow x = 7 \Rightarrow y = \dfrac{8 - x}{6} = \dfrac{8 - 7}{6} = \dfrac{1}{6}\)
\(\begin{cases} x - 6y = - 6 \\ x + 6y = 8 \end{cases}\ \Rightarrow 2x = 2 \Rightarrow x = 1 \Rightarrow y = \dfrac{8 - x}{6} = \dfrac{8 - 1}{6} = \dfrac{7}{6}\)

2) \(\begin{cases} y^{2} - 2xy = 32 \\ x^{2} + 6xy + 9y^{2} = 100 \end{cases}\ \Rightarrow \begin{cases} y^{2} - 2xy = 32 \\ (x + 3y)^{2} = 100 \end{cases}\ \Rightarrow \begin{cases} y^{2} - 2xy = 32 \\ x + 3y = \pm 10 \end{cases}\)
\(\begin{cases} y^{2} - 2xy = 32 \\ x + 3y = 10 \end{cases}\ \Rightarrow \begin{cases} y^{2} - 2y(10 - 3y) = 32 \\ x = 10 - 3y \end{cases}\)
\(y^{2} - 20y + 6y^{2} - 32 = 0\)
\(7y^{2} - 20y - 32 = 0\)
\(\dfrac{D}{4} = 100 + 7 \cdot 32 = 100 + 224 = 324 = 18^{2}\)
\(y_{1} = \dfrac{10 - 18}{7} = - \dfrac{8}{7};\)
\(x_{1} = 10 - 3 \cdot \left( - \dfrac{8}{7} \right) = 10 + \dfrac{24}{7} = \dfrac{70 + 24}{7} = \dfrac{94}{7}\)
\(y_{2} = \dfrac{10 + 18}{7} = \dfrac{28}{7} = 4;\)
\(x_{2} = 10 - 3 \cdot 4 = 10 - 12 = - 2\)
\(\begin{cases} y^{2} - 2xy = 32 \\ x + 3y = - 10 \end{cases}\ \Rightarrow \begin{cases} y^{2} - 2y( - 10 - 3y) = 32 \\ x = - 10 - 3y \end{cases}\)
\(y^{2} + 20y + 6y^{2} - 32 = 0\)
\(7y^{2} + 20y - 32 = 0\)
\(\dfrac{D}{4} = 18^{2}\)
\(y_{3} = \dfrac{- 10 - 18}{7} = - \dfrac{28}{7} = - 4;\)
\(x_{3} = - 10 - 3 \cdot ( - 4) = - 10 + 12 = 2\)
\(y_{4} = \dfrac{- 10 + 18}{7} = \dfrac{8}{7}\)
\(x_{4} = - 10 - 3 \cdot \dfrac{8}{7} = - 10 - \dfrac{24}{7} = \dfrac{- 70 - 24}{7} = - \dfrac{94}{7}\)

3) \(\begin{cases} x^{2} + y^{2} = 25 \\ xy = 12 \end{cases}\ \Rightarrow \begin{cases} x^{2} + \frac{144}{x^{2}} = 25 \\ y = \frac{12}{x} \end{cases}\ \Rightarrow \left| x^{2} = t \right| \Rightarrow \begin{cases} t + \frac{144}{t} = 25 \\ y = \frac{12}{x} \end{cases}\)
\(t^{2} + 144 = 25t\)
\(t^{2} - 25t + 144 = 0\)
\(D = 625 - 4 \cdot 144 = 625 - 576 = 49\)
\(t_{1} = \dfrac{25 - 7}{2} = \dfrac{18}{2} = 9 \Rightarrow x^{2} = 9 \Rightarrow x_{1{,}2} = \pm 3 \Rightarrow y_{1{,}2} = \pm 4\)
\(t_{2} = \dfrac{25 + 7}{2} = 16 \Rightarrow x^{2} = 16 \Rightarrow x_{3{,}4} = \pm 4 \Rightarrow y_{3{,}4} = \pm 3\)

4) \(\begin{cases} 9x^{2} + y^{2} = 10 \\ xy = - 1 \end{cases}\ \Rightarrow \begin{cases} 9x^{2} + \frac{1}{x^{2}} = 10 \\ y = - \frac{1}{x} \end{cases}\ \Rightarrow \left| x^{2} = t \right| \Rightarrow \begin{cases} 9t + \frac{1}{t} = 10 \\ y = - \frac{1}{x} \end{cases}\)
\(9t^{2} + 1 = 10t\)
\(9t^{2} - 10t + 1 = 0\)
\(\dfrac{D}{4} = 25 - 9 = 16\)
\(t_{1} = \dfrac{5 - 4}{9} = \dfrac{1}{9} \Rightarrow x^{2} = \dfrac{1}{9} \Rightarrow x_{1{,}2} = \pm \dfrac{1}{3} \Rightarrow y_{1{,}2} = \mp 3\)
\(t_{2} = \dfrac{5 + 4}{9} = 1 \Rightarrow x^{2} = 1 \Rightarrow x_{3{,}4} = \pm 1 \Rightarrow y_{3{,}4} = \mp 1\)

Ответ: 1) \(\left( 7;\frac{1}{6} \right);\left( 1;\frac{7}{6} \right)\); 2) \(\left( \frac{94}{7}; - \frac{8}{7} \right);\ ( - 2;\ 4);\ (2; - 4);\left( - \frac{94}{7};\frac{8}{7} \right)\); 3) \((3;\ 4);\ ( - 3; - 4);\ (4;\ 3);\ ( - 4; - 3)\); 4) \(\left( \frac{1}{3}; - 3 \right);\left( - \frac{1}{3};\ 3 \right);\ (1; - 1);\ ( - 1;\ 1)\).