Страница 129 номер 466, ГДЗ по алгебре за 9 класс к учебнику Мерзляка
1) \(\begin{cases} 3y - 2xy = 2 \\ x + 2xy = 5 \end{cases}\ \Rightarrow 3y - 2xy + x + 2xy = 2 + 5 \Rightarrow\)
\(\Rightarrow 3y + x = 7 \Rightarrow x = 7 - 3y\)
\(3y - 2xy = 2\)
\(3y - 2y(7 - 3y) = 2\)
\(3y - 14y + 6y^{2} = 2\)
\(6y^{2} - 11y - 2 = 0\)
\(D = 121 + 4 \cdot 6 \cdot 2 = 121 + 48 = 169\)
\(y_{1} = \dfrac{11 - 13}{12} = - \dfrac{2}{12} = - \dfrac{1}{6};\ x_{1} = 7 - 3 \cdot \left( - \dfrac{1}{6} \right) = 7 + \dfrac{1}{2} = 7{,}5\)
\(y_{2} = \dfrac{11 + 13}{12} = \dfrac{24}{12} = 2;\ x_{2} = 7 - 3 \cdot 2 = 1\)
2) \(\begin{cases} xy + y = 30 \\ xy + x = 28 \end{cases}\ \Rightarrow y - x = 2 \Rightarrow y = x + 2\)
\(xy + x = 28\)
\(x(x + 2) + x = 28\)
\(x^{2} + 2x + x - 28 = 0\)
\(x^{2} + 3x - 28 = 0\)
\(D = 9 + 4 \cdot 28 = 9 + 112 = 121\)
\(x_{1} = \dfrac{- 3 - 11}{2} = - \dfrac{14}{2} = - 7;\ y_{1} = - 7 + 2 = - 5\)
\(x_{2} = \dfrac{- 3 + 11}{2} = \dfrac{8}{2} = 4;\ y_{2} = 4 + 2 = 6\)
3) \(\begin{cases} x^{2} + y^{2} + x + y = 18 \\ x^{2} - y^{2} + x - y = 6 \end{cases}\)
Сложим оба уравнения
\(2x^{2} + 2x = 24\)
\(x^{2} + x - 12 = 0\)
\((x - 3)(x + 4) = 0\)
\(x_{1} = 3;\ x_{2} = - 4\)
Из первого вычтем другое
\(2y^{2} + 2y = 12\)
\(y^{2} + y - 6 = 0\)
\((y - 2)(y + 3) = 0\)
\(y_{1} = 2;\ y_{2} = - 3\)
Проверим, какие пары соответствуют решению по первому уравнению.
\((3;\ 2) \Rightarrow 3^{2} + 2^{2} + 3 + 2 = 9 + 4 + 5 = 18\)
\(( - 4;\ 2) \Rightarrow ( - 4)^{2} + 2^{2} - 4 + 2 = 16 + 4 - 2 = 18\)
\((3; - 3) \Rightarrow 3^{2} + ( - 3)^{2} + 3 - 3 = 9 + 9 = 18\)
\(( - 4; - 3) \Rightarrow ( - 4)^{2} + ( - 3)^{2} - 4 - 3 = 16 + 9 - 7 = 18\)
4) \(\begin{cases} 2x^{2} - 5xy + 3x - 2y = 10 \\ 5xy - 2x^{2} + 7x - 8y = 10 \end{cases}\)
Сложив оба уравнения, получим
\(10x - 10y = 20\)
\(x - y = 2\)
\(x = 2 + y\)
\(2x^{2} - 5xy + 3x - 2y = 10\)
\(2(2 + y)^{2} - 5y(2 + y) + 3(2 + y) - 2y = 10\)
\(8 + 8y + 2y^{2} - 10y - 5y^{2} + 6 + 3y - 2y - 10 = 0\)
\(- 3y^{2} - y + 4 = 0\)
\(D = 1 + 4 \cdot 3 \cdot 4 = 1 + 48 = 49\)
\(y_{1} = \dfrac{1 - 7}{- 6} = 1;\ x_{1} = 2 + 1 = 3\)
\(y_{2} = \dfrac{1 + 7}{- 6} = - \dfrac{8}{6} = - \dfrac{4}{3};\ x_{2} = 2 - \dfrac{4}{3} = \dfrac{6 - 4}{3} = \dfrac{2}{3}\)
Ответ: 1) \(\left( 7{,}5; - \frac{1}{6} \right);\ (1;\ 2)\); 2) \(( - 7; - 5);\ (4;\ 6)\); 3) \((3;\ 2);\ ( - 4;\ 2);\ (3; - 3);\ ( - 4; - 3)\); 4) \((3;\ 1);\left( \frac{2}{3}; - \frac{4}{3} \right)\).
