Страница 128 номер 461, ГДЗ по алгебре за 9 класс к учебнику Мерзляка

1) \(\begin{cases} \frac{1}{x} + \frac{1}{y} = \frac{3}{2} \\ x - y = 1 \end{cases}\ \Rightarrow \begin{cases} 2y + 2x = 3xy \\ x = y + 1 \end{cases}\)
\(2y + 2x = 3xy\)
\(2y + 2(y + 1) = 3y(y + 1)\)
\(2y + 2y + 2 = 3y^{2} + 3y\)
\(3y^{2} - y - 2 = 0\)
\(D = 1 + 4 \cdot 3 \cdot 2 = 1 + 24 = 25\)
\(y_{1} = \dfrac{1 - 5}{6} = - \dfrac{2}{3};\ x_{1} = - \dfrac{2}{3} + 1 = \dfrac{1}{3}\)
\(y_{2} = \dfrac{1 + 5}{6} = 1;\ x_{2} = 1 + 1 = 2\)

2) \(\begin{cases} \frac{1}{x} - \frac{1}{y} = \frac{4}{5} \\ 3x + y = 8 \end{cases}\ \Rightarrow \begin{cases} 5y - 5x = 4xy \\ y = 8 - 3x \end{cases}\)
\(5y - 5x = 4xy\)
\(5(8 - 3x) - 5x = 4x(8 - 3x)\)
\(40 - 15x - 5x = 32x - 12x^{2}\)
\(12x^{2} - 52x + 40 = 0\)
\(3x^{2} - 13x + 10 = 0\)
\(D = 169 - 4 \cdot 3 \cdot 10 = 169 - 120 = 49\)
\(x_{1} = \dfrac{13 - 7}{6} = 1;\ y_{1} = 8 - 3 = 5\)
\(x_{2} = \dfrac{13 + 7}{6} = \dfrac{20}{6} = \dfrac{10}{3};\ y_{2} = 8 - 3 \cdot \dfrac{10}{3} = - 2\)

Ответ: 1) \(\left( \frac{1}{3}; - \frac{2}{3} \right);\ (2;\ 1)\); 2) \((1;\ 5);\left( \frac{10}{3}; - 2 \right)\).