Страница 127 номер 455, ГДЗ по алгебре за 9 класс к учебнику Мерзляка

1) \(\begin{cases} 3x + 4y = 24 \\ xy = 12 \end{cases}\ \Rightarrow \begin{cases} 3x + 4 \cdot \frac{12}{x} = 24 \\ y = \frac{12}{x} \end{cases}\)
\(3x + \dfrac{48}{x} = 24\)
\(3x^{2} + 48 = 24x\)
\(3x^{2} - 24x + 48 = 0\)
\(x^{2} - 8x + 16 = 0\)
\((x - 4)^{2} = 0 \Rightarrow x = 4 \Rightarrow y = \dfrac{12}{4} = 3\)

2) \(\begin{cases} y + 2x = 0 \\ x^{2} + y^{2} - 6y = 0 \end{cases}\ \Rightarrow \begin{cases} y = - 2x \\ x^{2} + ( - 2x)^{2} - 6 \cdot ( - 2x) = 0 \end{cases}\ \Rightarrow\)
\(x^{2} + 4x^{2} + 12x = 0\)
\(5x^{2} + 12x = 0\)
\(5x\left( x + \dfrac{12}{5} \right) = 0\)
\(x_{1} = 0;\ y_{1} = - 2 \cdot 0 = 0\)
\(x_{2} = - \dfrac{12}{5};\ y_{2} = - 2 \cdot \left( - \dfrac{12}{5} \right) = \dfrac{24}{5}\)

3) \(\begin{cases} x^{2} - xy - y^{2} = 19 \\ x - y = 7 \end{cases}\ \Rightarrow \begin{cases} (y + 7)^{2} - y(y + 7) - y^{2} = 19 \\ x = y + 7 \end{cases}\)
\((y + 7)^{2} - y(y + 7) - y^{2} = 19\)
\(y^{2} + 14y + 49 - y^{2} - 7y - y^{2} - 19 = 0\)
\(- y^{2} + 7y + 30 = 0\)
\(D = 49 + 120 = 169\)
\(y_{1} = \dfrac{- 7 - 13}{- 2} = 10;\ x_{1} = 10 + 7 = 17\)
\(y_{2} = \dfrac{- 7 + 13}{- 2} = - 3;\ x_{2} = - 3 + 7 = 4\)

4) \(\begin{cases} x + y = 5 \\ (x - 3)(y + 5) = 6 \end{cases}\ \Rightarrow \begin{cases} y = 5 - x \\ (x - 3)(5 - x + 5) = 6 \end{cases}\)
\((x - 3)(10 - x) = 6\)
\((x - 3)(x - 10) = - 6\)
\(x^{2} - 13x + 30 + 6 = 0\)
\(x^{2} - 13x + 36 = 0\)
\(D = 169 - 4 \cdot 36 = 169 - 144 = 25\)
\(x_{1} = \dfrac{13 - 5}{2} = 4;\ y_{1} = 5 - 4 = 1\)
\(x_{2} = \dfrac{13 + 5}{2} = 9;\ y_{2} = 5 - 9 = - 4\)

5) \(\begin{cases} 4y - 3x = 4 \\ 5x^{2} + 16y = 60 \end{cases}\ \Rightarrow \begin{cases} y = \frac{3x + 4}{4} \\ 5x^{2} + 16 \cdot \frac{3x + 4}{4} = 60 \end{cases}\ \Rightarrow\)
\(5x^{2} + 4(3x + 4) = 60\)
\(5x^{2} + 12x + 16 - 60 = 0\)
\(5x^{2} + 12x - 44 = 0\)
\(\dfrac{D}{4} = 36 + 220 = 256 = 16^{2}\)
\(x_{1} = \dfrac{- 6 - 16}{5} = - \dfrac{22}{5} = - 4{,}4;\ y_{1} = \dfrac{3 \cdot ( - 4{,}4) + 4}{4} = - 2{,}3\)
\(x_{2} = \dfrac{- 6 + 16}{5} = 2;\ y_{2} = \dfrac{3 \cdot 2 + 4}{4} = 2{,}5\)

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

Ответ: 1) \((4;\ 3)\); 2) \((0;\ 0);\left( - \frac{12}{5};\frac{24}{5} \right)\); 3) \((17;\ 10);\ (4; - 3)\); 4) \((4;\ 1);\ (9; - 4)\); 5) \(( - 4{,}4; - 2{,}3);\ (2;\ 2{,}5)\); 6) \((4; - 1);\ (0;\ 3)\).