Страница 257 номер 984, ГДЗ по алгебре за 9 класс к учебнику Мерзляка

1) \(\begin{cases} x - 4y = - 6 \\ x^{2} + 4y^{2} = 8 \end{cases}\ \Rightarrow \left\{ \begin{aligned} & x = 4y - 6 \\ & (4y - 6)^{2} + 4y^{2} = 8 \end{aligned} \right.\)
\((4y - 6)^{2} + 4y^{2} = 8\)
\(16y^{2} - 48y + 36 + 4y^{2} = 8\)
\(20y^{2} - 48y + 28 = 0\)
\(5y^{2} - 12y + 7 = 0\)
\(\dfrac{D}{4} = 36 - 35 = 1\)
\(y_{1} = \dfrac{6 - 1}{5} = 1;\ x_{1} = 4 \cdot 1 - 6 = - 2\)
\(y_{2} = \dfrac{6 + 1}{5} = \dfrac{7}{5};\ x_{2} = 4 \cdot \dfrac{7}{5} - 6 = \dfrac{28 - 30}{5} = - \dfrac{2}{5}\)

2) \(\left\{ \begin{aligned} & 3x + y = - 2 \\ & 3x^{2} - 2xy = 28 \end{aligned} \right.\ \Rightarrow \left\{ \begin{aligned} & y = - 3x - 2 \\ & 3x^{2} - 2x( - 3x - 2) = 28 \end{aligned} \right.\ \Rightarrow\)
\(3x^{2} - 2x( - 3x - 2) = 28\)
\(3x^{2} + 6x^{2} + 4x - 28 = 0\)
\(9x^{2} + 4x - 28 = 0\)
\(\dfrac{D}{4} = 4 + 9 \cdot 28 = 4 + 252 = 256 = 16^{2}\)
\(x_{1} = \dfrac{- 2 - 16}{9} = - \dfrac{18}{9} = - 2;\ y_{1} = - 3 \cdot ( - 2) - 2 = 4\)
\(x_{2} = \dfrac{- 2 + 16}{9} = \dfrac{14}{9};\ y_{2} = - 3 \cdot \dfrac{14}{9} - 2 = - \dfrac{42 + 18}{9} = - \dfrac{20}{3}\)

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

4) \(\begin{cases} 3x - 2y = 18 \\ xy = - 12 \end{cases}\ \Rightarrow \left\{ \begin{aligned} & y = \frac{3x - 18}{2} \\ & x \cdot \frac{3x - 18}{2} = - 12 \end{aligned} \right.\)
\(x \cdot \dfrac{3x - 18}{2} = - 12\)
\(x(3x - 18) = - 24\)
\(3x^{2} - 18x + 24 = 0\)
\(x^{2} - 6x + 8 = 0\)
\((x - 2)(x - 4) = 0\)
\(x_{1} = 2;\ y_{1} = \dfrac{3 \cdot 2 - 18}{2} = \dfrac{- 12}{2} = - 6\)
\(x_{2} = 4;\ y_{2} = \dfrac{3 \cdot 4 - 18}{2} = - \dfrac{6}{2} = - 3\)

5) \(\begin{cases} x^{2} - y^{2} = 21 \\ x + y = - 3 \end{cases}\ \Rightarrow \begin{cases} (x - y)(x + y) = 21 \\ x + y = - 3 \end{cases}\ \Rightarrow\)
\(\Rightarrow \begin{cases} (x - y) \cdot ( - 3) = 21 \\ x + y = - 3 \end{cases}\ \Rightarrow \begin{cases} x - y = - 7 \\ x + y = - 3 \end{cases}\ \Rightarrow 2x = - 10 \Rightarrow x = - 5\)
\(x + y = - 3 \Rightarrow y = - 3 - x = - 3 + 5 = 2\)

6) \(\left\{ \begin{aligned} & x^{2} - xy + y^{2} = 7 \\ & x - y = 1 \end{aligned} \right.\ \Rightarrow \begin{cases} x^{2} - xy + y^{2} = 7 \\ (x - y)^{2} = 1 \end{cases}\ \Rightarrow\)
\(\Rightarrow \begin{cases} x^{2} - xy + y^{2} = 7 \\ x^{2} - 2xy + y^{2} = 1 \end{cases}\ \Rightarrow xy = 6\)
\(\begin{cases} xy = 6 \\ x - y = 1 \end{cases}\ \Rightarrow \left\{ \begin{aligned} & y(y + 1) = 6 \\ & x = y + 1 \end{aligned} \right.\)
\(y(y + 1) = 6\)
\(y^{2} + y - 6 = 0\)
\((y - 2)(y + 3) = 0\)
\(y_{1} = 2;\ x_{1} = 2 + 1 = 3\)
\(y_{2} = - 3;\ x_{2} = - 3 + 1 = - 2\)

7) \(\begin{cases} (x - 1)(y - 1) = - 2 \\ x + y = 1 \end{cases}\ \Rightarrow \left\{ \begin{aligned} & xy - y - x + 1 = - 2 \\ & x + y - 1 = 0 \end{aligned} \right.\ \Rightarrow\)
\(\Rightarrow \begin{cases} xy - (x + y - 1) = - 2 \\ x + y - 1 = 0 \end{cases}\ \Rightarrow \begin{cases} xy = - 2 \\ x = 1 - y \end{cases}\ \Rightarrow \begin{cases} y(1 - y) = - 2 \\ x = 1 - y \end{cases}\)
\(y - y^{2} = - 2\)
\(y^{2} - y - 2 = 0\)
\((y + 1)(y - 2) = 0\)
\(y_{1} = - 1;\ x_{1} = 1 - ( - 1) = 2\)
\(y_{2} = 2;\ x_{2} = 1 - 2 = - 1\)

8) \(\begin{cases} (x - 2)(y + 1) = 1 \\ x - y = 3 \end{cases}\ \Rightarrow \left\{ \begin{aligned} & (y + 3 - 2)(y + 1) = 1 \\ & x = y + 3 \end{aligned} \right.\ \Rightarrow\)
\((y + 3 - 2)(y + 1) = 1\)
\((y + 1)(y + 1) = 1\)
\((y + 1)^{2} = 1\)
\(y + 1 = \pm 1\)
\(y_{1} = 1 - 1 = 0;\ x_{1} = 0 + 3 = 3\)
\(y_{2} = - 1 - 1 = - 2;\ x_{2} = - 2 + 3 = 1\)

9) \(\begin{cases} \frac{1}{x} + \frac{1}{y} = \frac{3}{8} \\ x + y = 12 \end{cases}\ \Rightarrow \left\{ \begin{aligned} & 8(x + y) = 3xy \\ & x + y = 12 \end{aligned} \right.\ \Rightarrow \begin{cases} 8 \cdot 12 = 3xy \\ x + y = 12 \end{cases}\ \Rightarrow\)
\(\Rightarrow \left\{ \begin{aligned} & xy = 32 \\ & y = 12 - x \end{aligned} \right.\ \Rightarrow \begin{cases} x(12 - x) = 32 \\ y = 12 - x \end{cases}\)
\(x(12 - x) = 32\)
\(12x - x^{2} = 32\)
\(x^{2} - 12x + 32 = 0\)
\((x - 4)(x - 8) = 0\)
\(x_{1} = 4;\ y_{1} = 12 - 4 = 8\)
\(x_{2} = 8;\ y_{2} = 12 - 8 = 4\)

10) \(\begin{cases} \frac{1}{x} - \frac{1}{y} = \frac{4}{5} \\ y - x = 4 \end{cases}\ \Rightarrow \begin{cases} 5(y - x) = 4xy \\ y - x = 4 \end{cases}\ \Rightarrow \begin{cases} 5 \cdot 4 = 4xy \\ y - x = 4 \end{cases}\ \Rightarrow\)
\(\Rightarrow \left\{ \begin{aligned} & xy = 5 \\ & y = x + 4 \end{aligned} \right.\ \Rightarrow \begin{cases} x(x + 4) = 5 \\ y = x + 4 \end{cases}\)
\(x(x + 4) = 5\)
\(x^{2} + 4x - 5 = 0\)
\((x - 1)(x + 5) = 0\)
\(x_{1} = 1;\ y_{1} = 1 + 4 = 5\)
\(x_{2} = - 5;\ y_{2} = - 5 + 4 = - 1\)

Ответ: 1) \(( - 2;\ 1);\left( - \frac{2}{5};\frac{7}{5} \right)\); 2) \(( - 2;\ 4);\left( \frac{14}{9}; - \frac{20}{3} \right)\); 3) \(\left( 10;\frac{3}{2} \right);\ (3;\ 5)\); 4) \((2; - 6);\ (4; - 3)\); 5) \(( - 5;\ 2)\); 6) \((3;\ 2);\ ( - 2; - 3)\); 7) \(( - 1;\ 2);\ (2; - 1)\); 8) \((3;\ 0);\ (1; - 2)\); 9) \((4;\ 8);\ (8;\ 4)\); 10) \((1;\ 5);\ ( - 5; - 1)\).