Страница 257 номер 985, ГДЗ по алгебре за 9 класс к учебнику Мерзляка

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

Решения \((1;\ 2);\ ( - 1; - 2);\ (2;\ 1);\ ( - 2; - 1)\)

2) \(\begin{cases} xy + x + y = 11 \\ xy(x + y) = 30 \end{cases}\ \Rightarrow |t = xy;\ h = x + y| \Rightarrow \left\{ \begin{aligned} & t + h = 11 \\ & th = 30 \end{aligned} \right.\)

По теореме, обратной теореме Виета, \(t\) и \(h\) – корни уравнения \(z^{2} - 11z + 30 = 0\), то есть числа 5 и 6.

Так же найдём \(x\) и \(y\): при \(t = 5\) это корни уравнения \(z^{2} - 6z + 5 = 0\) (числа 1 и 5), при \(t = 6\) – уравнения \(z^{2} - 5z + 6 = 0\) (числа 2 и 3).

\(\left\{ \begin{aligned} & t = 5 \\ & h = 6 \end{aligned} \right.\ \Rightarrow \begin{cases} xy = 5 \\ x + y = 6 \end{cases}\ \Rightarrow \begin{cases} x_{1} = 1;\ y_{1} = 5 \\ x_{2} = 5;\ y_{2} = 1 \end{cases}\)
\(\begin{cases} t = 6 \\ h = 5 \end{cases}\ \Rightarrow \begin{cases} xy = 6 \\ x + y = 5 \end{cases}\ \Rightarrow \begin{cases} x_{3} = 2;\ y_{3} = 3 \\ x_{4} = 3;\ y_{4} = 2 \end{cases}\)

Решения \((1;\ 5);\ (5;\ 1);\ (2;\ 3);\ (3;\ 2)\)

3) \(\begin{cases} x + xy + y = 5 \\ x - xy + y = 1 \end{cases}\ \Rightarrow 2x + 2y = 6 \Rightarrow x + y = 3 \Rightarrow y = 3 - x\)
\(x + xy + y = 5\)
\(x + x(3 - x) + 3 - x = 5\)
\(x + 3x - x^{2} + 3 - x = 5\)
\(x^{2} - 3x + 2 = 0\)
\((x - 1)(x - 2) = 0\)
\(x_{1} = 1;\ y_{1} = 3 - 1 = 2\)
\(x_{2} = 2;\ y_{2} = 3 - 2 = 1\)

Решения \((1;\ 2);\ (2;\ 1)\)

4) \(\begin{cases} x - y = 2 \\ \frac{x}{y} - \frac{y}{x} = \frac{5}{6} \end{cases}\ \Rightarrow \left\{ \begin{aligned} & x = y + 2 \\ & 6\left( x^{2} - y^{2} \right) = 5xy \end{aligned} \right.\ \Rightarrow \begin{cases} x = y + 2 \\ 6(x - y)(x + y) = 5xy \end{cases}\ \Rightarrow\)
\(6(x - y)(x + y) = 5xy\)
\(6 \cdot 2 \cdot (x + y) = 5xy\)
\(12(y + 2 + y) = 5y(y + 2)\)
\(12(2y + 2) = 5y^{2} + 10y\)
\(24y + 24 = 5y^{2} + 10y\)
\(5y^{2} - 14y - 24 = 0\)
\(\dfrac{D}{4} = 49 + 5 \cdot 24 = 49 + 120 = 169 = 13^{2}\)
\(y_{1} = \dfrac{7 - 13}{5} = - \dfrac{6}{5};\ x_{1} = - \dfrac{6}{5} + 2 = \dfrac{- 6 + 10}{5} = \dfrac{4}{5}\)
\(y_{2} = \dfrac{7 + 13}{5} = 4;\ x_{2} = 4 + 2 = 6\)

Решения \(\left( \frac{4}{5}; - \frac{6}{5} \right);\ (6;\ 4)\)

5) \(\left\{ \begin{aligned} & x^{2} + 2xy + y^{2} = 25 \\ & x^{2} - 3xy = 4 \end{aligned} \right.\ \Rightarrow \begin{cases} (x + y)^{2} = 5^{2} \\ x^{2} - 3xy = 4 \end{cases}\)
\(\begin{cases} x + y = 5 \\ x^{2} - 3xy = 4 \end{cases}\ \Rightarrow \left\{ \begin{aligned} & y = 5 - x \\ & x^{2} - 3x(5 - x) = 4 \end{aligned} \right.\ \Rightarrow\)
\(x^{2} - 3x(5 - x) = 4\)
\(x^{2} - 15x + 3x^{2} = 4\)
\(4x^{2} - 15x - 4 = 0\)
\(D = 225 + 4^{3} = 225 + 64 = 289 = 17^{2}\)
\(x_{1} = \dfrac{15 - 17}{8} = - \dfrac{2}{8} = - \dfrac{1}{4};\ y_{1} = 5 - \left( - \dfrac{1}{4} \right) = 5\dfrac{1}{4}\)
\(x_{2} = \dfrac{15 + 17}{8} = 4;\ y_{2} = 5 - 4 = 1\)
\(\begin{cases} x + y = - 5 \\ x^{2} - 3xy = 4 \end{cases}\ \Rightarrow \begin{cases} y = - x - 5 \\ x^{2} + 3x(x + 5) = 4 \end{cases}\)
\(x^{2} + 3x(x + 5) = 4\)
\(x^{2} + 3x^{2} + 15x - 4 = 0\)
\(4x^{2} + 15x - 4 = 0\)
\(D = 17^{2}\)
\(x_{3} = \dfrac{- 15 - 17}{8} = - 4;\ y_{3} = - ( - 4) - 5 = - 1\)
\(x_{4} = \dfrac{- 15 + 17}{8} = \dfrac{1}{4};\ y_{4} = - \dfrac{1}{4} - 5 = - 5\dfrac{1}{4}\)

Решения \(\left( - \frac{1}{4};\ 5\frac{1}{4} \right);\ (4;\ 1);\ ( - 4; - 1);\left( \frac{1}{4}; - 5\frac{1}{4} \right)\)

6) \(\begin{cases} 2x^{2} - y^{2} = 14 \\ xy = - 6 \end{cases}\ \Rightarrow \left\{ \begin{aligned} & 2x^{2} - \frac{36}{x^{2}} = 14 \\ & y = - \frac{6}{x} \end{aligned} \right.\ \Rightarrow \left| t = x^{2} \right| \Rightarrow \begin{cases} 2t - \frac{36}{t} = 14 \\ y = - \frac{6}{x} \end{cases}\)
\(2t^{2} - 36 = 14t\)
\(2t^{2} - 14t - 36 = 0\)
\(t^{2} - 7t - 18 = 0\)
\((t + 2)(t - 9) = 0\)
\(t = - 2 < 0\)
\(t = 9 \Rightarrow x^{2} = 9 \Rightarrow \begin{cases} x_{1} = 3;\ y_{1} = - \frac{6}{3} = - 2 \\ x_{2} = - 3;\ y_{2} = - \frac{6}{- 3} = 2 \end{cases}\)

Решения \((3; - 2);\ ( - 3;\ 2)\)

7) \(\begin{cases} \frac{1}{x} - \frac{1}{y} = \frac{1}{10} \\ xy = 50 \end{cases}\ \Rightarrow \begin{cases} 10(y - x) = xy \\ xy = 50 \end{cases}\ \Rightarrow \begin{cases} 10(y - x) = 50 \\ xy = 50 \end{cases}\ \Rightarrow \begin{cases} y - x = 5 \\ xy = 50 \end{cases}\)
\(\Rightarrow \begin{cases} y = 5 + x \\ x(5 + x) = 50 \end{cases}\)
\(5x + x^{2} - 50 = 0\)
\(x^{2} + 5x - 50 = 0\)
\((x - 5)(x + 10) = 0\)
\(x_{1} = 5;\ y_{1} = 10\)
\(x_{2} = - 10;\ y_{2} = - 5\)

Решения \((5;\ 10);\ ( - 10; - 5)\)

8) \(\begin{cases} \frac{x}{y} - \frac{y}{x} = \frac{16}{15} \\ x^{2} - y^{2} = 16 \end{cases}\ \Rightarrow \begin{cases} 15\left( x^{2} - y^{2} \right) = 16xy \\ x^{2} - y^{2} = 16 \end{cases}\ \Rightarrow \begin{cases} 15 \cdot 16 = 16xy \\ x^{2} - y^{2} = 16 \end{cases}\ \Rightarrow\)
\(\Rightarrow \begin{cases} xy = 15 \\ x^{2} - y^{2} = 16 \end{cases}\ \Rightarrow \left\{ \begin{aligned} & y = \dfrac{15}{x} \\ & x^{2} - \dfrac{225}{x^{2}} = 16 \end{aligned} \right.\ \Rightarrow \left| t = x^{2} \right| \Rightarrow \begin{cases} y = \frac{15}{x} \\ t - \frac{225}{t} = 16 \end{cases}\)
\(t^{2} - 225 = 16t\)
\(t^{2} - 16t - 225 = 0\)
\(\dfrac{D}{4} = 64 + 225 = 289 = 17^{2}\)
\(t_{1} = 8 - 17 = - 9 < 0\)
\(t_{2} = 8 + 17 = 25 \Rightarrow x^{2} = 25 \Rightarrow \left\{ \begin{aligned} & x_{1} = - 5;\ y_{1} = - 3 \\ & x_{2} = 5;\ y_{2} = 3 \end{aligned} \right.\)

Решения \(( - 5; - 3);\ (5;\ 3)\)

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

Решения \((3;\ 4);\ ( - 3; - 4);\ (4;\ 3);\ ( - 4; - 3)\)

10) \(\begin{cases} x^{2} + 2xy = 5 \\ y^{2} - 4xy = - 4 \end{cases}\ \Rightarrow x^{2} + 2xy + y^{2} - 4xy = 5 - 4 \Rightarrow\)
\(\Rightarrow x^{2} - 2xy + y^{2} = 1 \Rightarrow\)
\(\Rightarrow (x - y)^{2} = 1 \Rightarrow x - y = 1\ \text{или}\ x - y = - 1 \Rightarrow y = x - 1\ \text{или}\ y = x + 1\)
\(\left\{ \begin{aligned} & y = x - 1 \\ & x^{2} + 2x(x - 1) = 5 \end{aligned} \right.\ \Rightarrow \begin{cases} y = x - 1 \\ x^{2} + 2x^{2} - 2x - 5 = 0 \end{cases}\)
\(3x^{2} - 2x - 5 = 0\)
\(\dfrac{D}{4} = 1 + 15 = 16\)
\(x_{1} = \dfrac{1 - 4}{3} = - 1;\ y_{1} = - 1 - 1 = - 2\)
\(x_{2} = \dfrac{1 + 4}{3} = \dfrac{5}{3};\ y_{2} = \dfrac{5}{3} - 1 = \dfrac{2}{3}\)
\(\left\{ \begin{aligned} & y = x + 1 \\ & x^{2} + 2x(x + 1) = 5 \end{aligned} \right.\ \Rightarrow \begin{cases} y = x + 1 \\ x^{2} + 2x^{2} + 2x - 5 = 0 \end{cases}\)
\(3x^{2} + 2x - 5 = 0\)
\(\dfrac{D}{4} = 16\)
\(x_{3} = \dfrac{- 1 - 4}{3} = - \dfrac{5}{3};\ y_{3} = - \dfrac{5}{3} + 1 = - \dfrac{2}{3}\)
\(x_{4} = \dfrac{- 1 + 4}{3} = 1;\ y_{4} = 1 + 1 = 2\)

Решения \(( - 1; - 2);\left( \frac{5}{3};\frac{2}{3} \right);\left( - \frac{5}{3}; - \frac{2}{3} \right);\ (1;\ 2)\)

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