Страница 126 номер 451, ГДЗ по алгебре за 9 класс к учебнику Мерзляка

1) \(\begin{cases} y = x + 3 \\ x^{2} - 2y = 9 \end{cases}\ \Rightarrow \begin{cases} y = x + 3 \\ x^{2} - 2(x + 3) = 9 \end{cases}\)
\(x^{2} - 2x - 6 - 9 = 0\)
\(x^{2} - 2x - 15 = 0\)
\(\dfrac{D}{4} = 1 + 15 = 16\)
\(x_{1} = 1 - 4 = - 3;\ y_{1} = - 3 + 3 = 0\)
\(x_{2} = 1 + 4 = 5;\ y_{2} = 5 + 3 = 8\)

2) \(\begin{cases} x + y = 5 \\ xy = 4 \end{cases}\ \Rightarrow \begin{cases} y = 5 - x \\ x(5 - x) = 4 \end{cases}\ \Rightarrow \begin{cases} y = 5 - x \\ 5x - x^{2} = 4 \end{cases}\)
\(5x - x^{2} = 4\)
\(x^{2} - 5x + 4 = 0\)
\(D = 25 - 4 \cdot 4 = 9\)
\(x_{1} = \dfrac{5 - 3}{2} = 1;\ y_{1} = 5 - 1 = 4\)
\(x_{2} = \dfrac{5 + 3}{2} = 4;\ y_{2} = 5 - 4 = 1\)

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

4) \(\begin{cases} x - 4y = 2 \\ xy + 2y = 8 \end{cases}\ \Rightarrow \begin{cases} x = 4y + 2 \\ y(4y + 2) + 2y = 8 \end{cases}\)
\(y(4y + 2) + 2y = 8\)
\(4y^{2} + 2y + 2y - 8 = 0\)
\(4y^{2} + 4y - 8 = 0\)
\(y^{2} + y - 2 = 0\)
\(D = 1 + 4 \cdot 2 = 9\)
\(y_{1} = \dfrac{- 1 - 3}{2} = - 2;\ x_{1} = 4 \cdot ( - 2) + 2 = - 6\)
\(y_{2} = \dfrac{- 1 + 3}{2} = 1;\ x_{2} = 4 \cdot 1 + 2 = 6\)

5) \(\begin{cases} xy = 15 \\ 2x - y = 7 \end{cases}\ \Rightarrow \begin{cases} x(2x - 7) = 15 \\ y = 2x - 7 \end{cases}\)
\(x(2x - 7) = 15\)
\(2x^{2} - 7x - 15 = 0\)
\(D = 49 + 4 \cdot 2 \cdot 15 = 49 + 120 = 169\)
\(x_{1} = \dfrac{7 - 13}{4} = - \dfrac{6}{4} = - \dfrac{3}{2};\ y_{1} = 2 \cdot \left( - \dfrac{3}{2} \right) - 7 = - 10\)
\(x_{2} = \dfrac{7 + 13}{4} = 5;\ y_{2} = 2 \cdot 5 - 7 = 3\)

6) \(\begin{cases} x - y = 4 \\ x^{2} + y^{2} = 8 \end{cases}\ \Rightarrow \begin{cases} x = 4 + y \\ (4 + y)^{2} + y^{2} = 8 \end{cases}\)
\((4 + y)^{2} + y^{2} = 8\)
\(16 + 8y + y^{2} + y^{2} - 8 = 0\)
\(2y^{2} + 8y + 8 = 0\)
\(y^{2} + 4y + 4 = 0\)
\((y + 2)^{2} = 0\)
\(y = - 2;\ x = 4 - 2 = 2\)

Ответ: 1) \(( - 3;\ 0);\ (5;\ 8)\); 2) \((1;\ 4);\ (4;\ 1)\); 3) \(( - 3; - 1);\ ( - 1;\ 1)\); 4) \(( - 6; - 2);\ (6;\ 1)\); 5) \(\left( - \frac{3}{2}; - 10 \right),(5;\ 3)\); 6) \((2; - 2)\).