Страница 128 номер 462, ГДЗ по алгебре за 9 класс к учебнику Мерзляка

1) \(\begin{cases} x + y - xy = 1 \\ xy(x + y) = 20 \end{cases}\ \Rightarrow \begin{cases} t = xy \\ h = x + y \end{cases}\ \Rightarrow \begin{cases} h - t = 1 \\ h \cdot t = 20 \end{cases}\ \Rightarrow \begin{cases} h = 1 + t \\ t(1 + t) = 20 \end{cases}\)
\(t(1 + t) = 20\)
\(t^{2} + t - 20 = 0\)
\((t - 4)(t + 5) = 0\)
\(t_{1} = 4;\ h_{1} = 1 + 4 = 5 \Rightarrow \begin{cases} xy = 4 \\ x + y = 5 \end{cases}\ \Rightarrow \begin{cases} x = 1 \\ y = 4 \end{cases}\ \ \text{или}\ \begin{cases} x = 4 \\ y = 1 \end{cases}\)
\(t_{2} = - 5;\ h_{2} = 1 - 5 = - 4 \Rightarrow\)
\(\Rightarrow \begin{cases} xy = - 5 \\ x + y = - 4 \end{cases}\ \Rightarrow \begin{cases} x = 1 \\ y = - 5 \end{cases}\ \ \text{или}\ \begin{cases} x = - 5 \\ y = 1 \end{cases}\)

2) \(\begin{cases} \frac{y}{x} - \frac{x}{y} = \frac{21}{10} \\ x + y = 3 \end{cases}\ \Rightarrow \begin{cases} 10y^{2} - 10x^{2} = 21xy \\ x = 3 - y \end{cases}\)
\(10y^{2} - 10x^{2} = 21xy\)
\(10y^{2} - 10(3 - y)^{2} = 21y(3 - y)\)
\(10y^{2} - 90 + 60y - 10y^{2} = 63y - 21y^{2}\)
\(21y^{2} - 3y - 90 = 0\)
\(7y^{2} - y - 30 = 0\)
\(D = 1 + 4 \cdot 7 \cdot 30 = 1 + 840 = 841 = 29^{2}\)
\(y_{1} = \dfrac{1 - 29}{14} = - \dfrac{28}{14} = - 2;\ x_{1} = 3 - ( - 2) = 5\)
\(y_{2} = \dfrac{1 + 29}{14} = \dfrac{30}{14} = \dfrac{15}{7};\ x_{2} = 3 - \dfrac{15}{7} = \dfrac{21 - 15}{7} = \dfrac{6}{7}\)

3) \(\begin{cases} \frac{x}{y} + \frac{6y}{x} = 5 \\ x^{2} + 4xy - 3y^{2} = 18 \end{cases}\)

Пусть \(\frac{x}{y} = t\), тогда \(\frac{x}{y} + \frac{6y}{x} = 5 \Rightarrow t + \frac{6}{t} = 5 \Rightarrow t^{2} + 6 = 5t \Rightarrow\)

\(\Rightarrow t^{2} - 5t + 6 = 0 \Rightarrow (t - 2)(t - 3) = 0 \Rightarrow t_{1} = 2;\ t_{2} = 3\).

Имеем две системы уравнений

\(\begin{cases} \frac{x}{y} = 2 \\ x^{2} + 4xy - 3y^{2} = 18 \end{cases}\ \Rightarrow \begin{cases} x = 2y \\ (2y)^{2} + 4y(2y) - 3y^{2} = 18 \end{cases}\ \Rightarrow\)
\(4y^{2} + 8y^{2} - 3y^{2} = 18\)
\(9y^{2} = 18\)
\(y_{1{,}2} = \pm \sqrt{2} \Rightarrow x_{1{,}2} = \pm 2\sqrt{2}\)
\(\begin{cases} \frac{x}{y} = 3 \\ x^{2} + 4xy - 3y^{2} = 18 \end{cases}\ \Rightarrow \begin{cases} x = 3y \\ (3y)^{2} + 4y(3y) - 3y^{2} = 18 \end{cases}\ \Rightarrow\)
\(9y^{2} + 12y^{2} - 3y^{2} = 18\)
\(18y^{2} = 18\)
\(y_{3{,}4} = \pm 1 \Rightarrow x_{3{,}4} = \pm 3\)

4) \(\begin{cases} \frac{1}{x} + \frac{1}{y} = \frac{5}{6} \\ \frac{1}{x} - \frac{1}{y} = \frac{1}{6} \end{cases}\ \Rightarrow \left| \frac{1}{x} = t;\frac{1}{y} = k \right| \Rightarrow \begin{cases} t + k = \frac{5}{6} \\ t - k = \frac{1}{6} \end{cases}\ \Rightarrow\)
\(\Rightarrow 2t = 1 \Rightarrow t = \dfrac{1}{2} \Rightarrow x = \dfrac{1}{t} \Rightarrow x = 2\)
\(t + k = \dfrac{5}{6} \Rightarrow k = \dfrac{5}{6} - t = \dfrac{5}{6} - \dfrac{1}{2} = \dfrac{5 - 3}{6} = \dfrac{2}{6} = \dfrac{1}{3} \Rightarrow y = \dfrac{1}{k} = 3\)

5) \(\begin{cases} \frac{y}{x} + xy = - 10 \\ \frac{5y}{x} - 2xy = 13 \end{cases}\ \Rightarrow \left| \frac{y}{x} = t;\ xy = k \right| \Rightarrow \begin{cases} t + k = - 10 \\ 5t - 2k = 13 \end{cases}\ \Rightarrow\)
\(\Rightarrow \begin{cases} 2t + 2k = - 20 \\ 5t - 2k = 13 \end{cases}\ \Rightarrow 7t = - 7 \Rightarrow t = - 1\)
\(t + k = - 10 \Rightarrow k = - 10 - t = - 10 - ( - 1) = - 9\)
\(\begin{cases} \frac{y}{x} = - 1 \\ xy = - 9 \end{cases}\ \Rightarrow \begin{cases} y = - x \\ x \cdot ( - x) = - 9 \end{cases}\ \Rightarrow \begin{cases} y = - x \\ - x^{2} = - 9 \end{cases}\ \Rightarrow\)
\(\Rightarrow \begin{cases} y = - x \\ x^{2} = 9 \end{cases}\ \Rightarrow \begin{cases} y_{1{,}2} = \mp 3 \\ x_{1{,}2} = \pm 3\ \end{cases}\)

6) \(\begin{cases} x^{2}y^{2} + xy = 6 \\ 2x - y = 3 \end{cases}\)
\(x^{2}y^{2} + xy = 6 \Rightarrow |xy = t| \Rightarrow t^{2} + t - 6 = 0 \Rightarrow\)
\(\Rightarrow (t - 2)(t + 3) = 0\) \[\begin{cases} x^{2}y^{2} + xy = 6 \\ 2x - y = 3 \end{cases}\ \Rightarrow \left[ \begin{array}{l} \begin{cases} xy = 2 \\ y = 2x - 3 \end{cases} \\ \begin{cases} xy = - 3 \\ y = 2x - 3 \end{cases} \end{array} \right.\ \Rightarrow \left[ \begin{array}{l} \begin{cases} x(2x - 3) = 2 \\ y = 2x - 3 \end{cases} \\ \begin{cases} x(2x - 3) = - 3 \\ y = 2x - 3 \end{cases} \end{array} \right.\ \Rightarrow\] \(x(2x - 3) = 2\)
\(2x^{2} - 3x - 2 = 0\)
\(D = 9 + 4 \cdot 2 \cdot 2 = 25\)
\(x_{1} = \dfrac{3 - 5}{4} = - \dfrac{1}{2};\ y_{1} = 2 \cdot \left( - \dfrac{1}{2} \right) - 3 = - 4\)
\(x_{2} = \dfrac{3 + 5}{4} = 2;\ y_{2} = 2 \cdot 2 - 3 = 1\)
\(x(2x - 3) = - 3\)
\(2x^{2} - 3x + 3 = 0\)
\(D = 9 - 4 \cdot 2 \cdot 3 < 0\) нет корней

7) \(\begin{cases} 3(x + y)^{2} + 2(x - 2y)^{2} = 5 \\ 2(x - 2y) - x - y = 1 \end{cases}\ \Rightarrow |x + y = t;\ x - 2y = k| \Rightarrow\)
\(\Rightarrow \begin{cases} 3t^{2} + 2k^{2} = 5 \\ 2k - t = 1 \end{cases}\ \Rightarrow \begin{cases} 3(2k - 1)^{2} + 2k^{2} = 5 \\ t = 2k - 1 \end{cases}\)
\(3(2k - 1)^{2} + 2k^{2} = 5\)
\(3\left( 4k^{2} - 4k + 1 \right) + 2k^{2} - 5 = 0\)
\(12k^{2} - 12k + 3 + 2k^{2} - 5 = 0\)
\(14k^{2} - 12k - 2 = 0\)
\(7k^{2} - 6k - 1 = 0\)
\(\dfrac{D}{4} = 9 + 7 = 16\)
\(k_{1} = \dfrac{3 - 4}{7} = - \dfrac{1}{7};\ t_{1} = 2 \cdot \left( - \dfrac{1}{7} \right) - 1 = - \dfrac{2}{7} - 1 = - \dfrac{9}{7}\)
\(k_{2} = \dfrac{3 + 4}{7} = 1;\ t_{2} = 2 \cdot 1 - 1 = 1\)
\(\begin{cases} x + y = - \frac{9}{7} \\ x - 2y = - \frac{1}{7} \end{cases}\ \Rightarrow x + y - x + 2y = - \dfrac{9}{7} + \dfrac{1}{7} \Rightarrow\)
\(\Rightarrow 3y = - \dfrac{8}{7} \Rightarrow y = - \dfrac{8}{21} \Rightarrow x = - \dfrac{9}{7} + \dfrac{8}{21} = \dfrac{- 27 + 8}{21} = - \dfrac{19}{21}\)
\(\begin{cases} x + y = 1 \\ x - 2y = 1 \end{cases}\ \Rightarrow 3y = 0 \Rightarrow y = 0 \Rightarrow x = 1 - 0 = 1\)

Ответ: 1) \((1;\ 4);\ (4;\ 1);\ (1; - 5);\ ( - 5;\ 1)\); 2) \((5; - 2);\left( \frac{6}{7};\frac{15}{7} \right)\); 3) \(\left( 2\sqrt{2};\sqrt{2} \right);\left( - 2\sqrt{2}; - \sqrt{2} \right);\ (3;\ 1);\ ( - 3; - 1)\); 4) \((2;\ 3)\); 5) \((3; - 3);\ ( - 3;\ 3)\); 6) \(\left( - \frac{1}{2}; - 4 \right);\ (2;\ 1)\); 7) \((1;\ 0);\left( - \frac{19}{21}; - \frac{8}{21} \right)\).