Страница 127 номер 456, ГДЗ по алгебре за 9 класс к учебнику Мерзляка

1) \(\begin{cases} x^{2} - xy + y^{2} = 63 \\ y - x = 3 \end{cases}\ \Rightarrow \begin{cases} x^{2} - 2xy + y^{2} + xy = 63 \\ x - y = - 3 \end{cases}\ \Rightarrow\)
\(\Rightarrow \begin{cases} (x - y)^{2} + xy = 63 \\ x = y - 3 \end{cases}\)
\((x - y)^{2} + xy = 63\)
\(( - 3)^{2} + y(y - 3) = 63\)
\(y^{2} - 3y + 9 - 63 = 0\)
\(y^{2} - 3y - 54 = 0\)
\(D = 9 + 4 \cdot 54 = 9 + 216 = 225 = 15^{2}\)
\(y_{1} = \dfrac{3 - 15}{2} = - 6;\ x_{1} = - 6 - 3 = - 9\)
\(y_{2} = \dfrac{3 + 15}{2} = 9;\ x_{2} = 9 - 3 = 6\)

2) \(\begin{cases} x + 2y = 1 \\ x^{2} + xy + 2y^{2} = 1 \end{cases}\ \Rightarrow \begin{cases} x = 1 - 2y \\ (1 - 2y)^{2} + y(1 - 2y) + 2y^{2} = 1 \end{cases}\ \Rightarrow\)
\((1 - 2y)^{2} + y(1 - 2y) + 2y^{2} = 1\)
\(1 - 4y + 4y^{2} + y - 2y^{2} + 2y^{2} - 1 = 0\)
\(4y^{2} - 3y = 0\)
\(4y\left( y - \dfrac{3}{4} \right) = 0\)
\(y_{1} = 0;\ x_{1} = 1 - 2 \cdot 0 = 1\)
\(y_{2} = \dfrac{3}{4};\ x_{2} = 1 - 2 \cdot \dfrac{3}{4} = 1 - \dfrac{3}{2} = - \dfrac{1}{2}\)

3) \(\begin{cases} (x - 1)(y - 2) = 2 \\ x + y = 6 \end{cases}\ \Rightarrow \begin{cases} (6 - y - 1)(y - 2) = 2 \\ x = 6 - y \end{cases}\)
\((5 - y)(y - 2) = 2\)
\(5y - y^{2} - 10 + 2y - 2 = 0\)
\(- y^{2} + 7y - 12 = 0\)
\(y^{2} - 7y + 12 = 0\)
\(D = 49 - 4 \cdot 12 = 1\)
\(y_{1} = \dfrac{7 - 1}{2} = 3;\ x_{1} = 6 - 3 = 3\)
\(y_{2} = \dfrac{7 + 1}{2} = 4;\ x_{2} = 6 - 4 = 2\)

4) \(\begin{cases} 5x - 2y = 3 \\ 3x^{2} - 8y = - 5 \end{cases}\ \Rightarrow \begin{cases} y = \frac{5x - 3}{2} \\ 3x^{2} - 8 \cdot \frac{5x - 3}{2} = - 5 \end{cases}\)
\(3x^{2} - 4(5x - 3) = - 5\)
\(3x^{2} - 20x + 12 + 5 = 0\)
\(3x^{2} - 20x + 17 = 0\)
\(\dfrac{D}{4} = 100 - 3 \cdot 17 = 100 - 51 = 49\)
\(x_{1} = \dfrac{10 - 7}{3} = 1;\ y_{1} = \dfrac{5 \cdot 1 - 3}{2} = 1\)
\(x_{2} = \dfrac{10 + 7}{3} = \dfrac{17}{3};\ y_{2} = \dfrac{1}{2}\left( 5 \cdot \dfrac{17}{3} - 3 \right) = \dfrac{85 - 9}{6} = \dfrac{76}{6} = \dfrac{38}{3}\)

Ответ: 1) \(( - 9; - 6);\ (6;\ 9)\); 2) \((1;\ 0);\left( - \frac{1}{2};\frac{3}{4} \right)\); 3) \((3;\ 3);\ (2;\ 4)\); 4) \(\left( \frac{17}{3};\frac{38}{3} \right);\ (1;\ 1)\).