《2025年初中升高中衔接读本南京出版社数学》

例5 解方程组：$\left\{\begin{array}{l} \frac{(x + 1)^2}{9} - \frac{(y - 1)^2}{4} = 1,\\ x - y = 1.\end{array} \right.$
【解答】原方程组可化为$\left\{\begin{array}{l} 4(x + 1)^2 - 9(y - 1)^2 = 36,\\ (x + 1) - (y - 1) = 3\end{array} \right.$
令$x + 1 = m$,$y - 1 = n$,则原方程组可化为$\left\{\begin{array}{l} 4m^2 - 9n^2 = 36,\\ m - n = 3,\end{array} \right.解得\left\{\begin{array}{l} m_1 = 3,\\ n_1 = 0;\end{array} \right.\left\{\begin{array}{l} m_2 = \frac{39}{5},\\ n_2 = \frac{24}{5}.\end{array} \right.$
代入所设,得原方程组的解为$\left\{\begin{array}{l} x_1 = 2,\\ y_1 = 1;\end{array} \right.\left\{\begin{array}{l} x_2 = \frac{34}{5},\\ y_2 = \frac{29}{5}.\end{array} \right.$

答案： 解：原方程组可化为
$\left\{\begin{array}{l}4(x + 1)^2 - 9(y - 1)^2 = 36, \\(x + 1) - (y - 1) = 3\end{array}\right.$
令$x + 1 = m$，$y - 1 = n$，则方程组化为
$\left\{\begin{array}{l}4m^2 - 9n^2 = 36, \\m - n = 3\end{array}\right.$
由$m - n = 3$得$m = n + 3$，代入$4m^2 - 9n^2 = 36$，得$4(n + 3)^2 - 9n^2 = 36$，整理得$5n^2 - 24n = 0$，解得$n_1 = 0$，$n_2 = \frac{24}{5}$。
当$n_1 = 0$时，$m_1 = 0 + 3 = 3$；当$n_2 = \frac{24}{5}$时，$m_2 = \frac{24}{5} + 3 = \frac{39}{5}$。
代入所设，得
$\left\{\begin{array}{l}x_1 = 3 - 1 = 2, \\y_1 = 0 + 1 = 1\end{array}\right.$
$\left\{\begin{array}{l}x_2 = \frac{39}{5} - 1 = \frac{34}{5}, \\y_2 = \frac{24}{5} + 1 = \frac{29}{5}\end{array}\right.$
原方程组的解为
$\left\{\begin{array}{l}x_1 = 2, \\y_1 = 1\end{array}\right.$
$\left\{\begin{array}{l}x_2 = \frac{34}{5}, \\y_2 = \frac{29}{5}\end{array}\right.$

1. 下列方程组是二元二次方程组的是( )

A.$\left\{\begin{array}{l} x + 3y = 5,\\ 4x - 6y = 9\end{array} \right.$
B.$\left\{\begin{array}{l} x^2 - 3y^2 = 2x + 5,\\ x + \frac{1}{y} = 3\end{array} \right.$
C.$\left\{\begin{array}{l} y = 3 - 5x^2,\\ x^2 - x - 6 = 0\end{array} \right.$
D.$\left\{\begin{array}{l} \sqrt{x^2 + 1} = 2y,\\ 3y = -x\end{array} \right.$

答案： C

2. 方程组$\left\{\begin{array}{l} x - y = 1,\\ x^2 - y^2 = 3\end{array} \right.$的解是( )

A.$\left\{\begin{array}{l} x = 2,\\ y = 1\end{array} \right.$
B.$\left\{\begin{array}{l} x = -1,\\ y = -2\end{array} \right.$
C.$\left\{\begin{array}{l} x = 3,\\ y = 2\end{array} \right.$
D.$\left\{\begin{array}{l} x = 1,\\ y = 2\end{array} \right.$

答案： A

3. 方程组$\left\{\begin{array}{l} x + y = 0,\\ 2x^2 + x + y - 3 = 0\end{array} \right.$的解的情况是( )

A.有两组相同的实数解
B.有两组不同的实数解
C.没有实数解
D.不能确定

答案： B

4. 方程组$\left\{\begin{array}{l} y = x^2,\\ y = x + m\end{array} \right.$有两组不同的实数解,则( )

A.$m \geq -\frac{1}{4}$
B.$m ＞ -\frac{1}{4}$
C.$-\frac{1}{4} ＜ m ＜ \frac{1}{4}$
D.以上答案都不对

答案： B

5. 方程组$\left\{\begin{array}{l} \frac{1}{x} + \frac{1}{y} = \frac{5}{6},\\ \frac{1}{x} \cdot \frac{1}{y} = \frac{1}{6}\end{array} \right.$的解是______.

答案： $\left\{\begin{array}{l} x_{1}=2,\ x_{2}=3,\\ y_{1}=3;\ y_{2}=2\end{array}\right. $

6. 请你写出一个以$\left\{\begin{array}{l} x = 1,\\ y = 1\end{array} \right.和\left\{\begin{array}{l} x = -1,\\ y = -1\end{array} \right.$为解的二元二次方程组,这个方程组可以是______.

答案： $\left\{\begin{array}{l} x-y=0,\\ x^{2}+y^{2}=2\end{array}\right. $

7. 把方程组$\left\{\begin{array}{l} x^2 + y^2 = 5,\\ x^2 - 5xy + 6y^2 = 0\end{array} \right.$化成两个二元二次方程组是______,______.

答案： $\left\{\begin{array}{l} x^{2}+y^{2}=5,\\ x-2y=0\end{array}\right. $$\left\{\begin{array}{l} x^{2}+y^{2}=5,\\ x-3y=0\end{array}\right. $

8. 若二元二次方程组$\left\{\begin{array}{l} x^2 - y^2 = 1,\\ y = k(x - 2) + 1\end{array} \right.$有唯一解,则实数$k = $______.

答案： $\pm 1$

9. 要使方程组$\left\{\begin{array}{l} x^2 + 2ay = 5,\\ y - x = 6a\end{array} \right.$有正整数解,则a的值是______.

答案： $\frac {1}{2},\frac {1}{6}$

10. 解下列方程组：
(1)$\left\{\begin{array}{l} x^2 + y^2 = 26,\\ xy = 5;\end{array} \right.$
(2)$\left\{\begin{array}{l} x^2 - 2xy + y^2 = 1,\\ 2x^2 - 5xy - 3y^2 = 0;\end{array} \right.$
(3)$\left\{\begin{array}{l} x^2 + y^2 = 10,\\ x^2 - 3xy + 2y^2 = 0;\end{array} \right.$
(4)$\left\{\begin{array}{l} (3x + 4y - 3)(3x + 4y + 3) = 0,\\ 3x + 2y = 5;\end{array} \right.$
(5)$\left\{\begin{array}{l} (x + y)(x + y - 1) = 0,\\ (x - y)(x - y - 1) = 0;\end{array} \right.$
(6)$\left\{\begin{array}{l} x^2 + y^2 = 3,\\ x^2 - y^2 = 0;\end{array} \right.$
(7)$\left\{\begin{array}{l} x + 2y = 3,\\ x^2 - 2y + 3x - 2 = 0;\end{array} \right.$
(8)$\left\{\begin{array}{l} 2x - 3y = 1,\\ 2x^2 - 3xy + y^2 - 4x + 3y - 3 = 0.\end{array} \right.$

答案： （1）$\left\{\begin{array}{l} x_{1}=5,\ x_{2}=1,\ x_{3}=-1,\ x_{4}=-5,\\ y_{1}=1;\ y_{2}=5;\ y_{3}=-5;\ y_{4}=-1\end{array}\right. $
（2）$\left\{\begin{array}{l} x_{1}=\frac {3}{2},\ x_{2}=\frac {1}{3},\ x_{3}=-\frac {1}{3},\ x_{4}=-\frac {3}{2},\\ y_{1}=\frac {1}{2};\ y_{2}=-\frac {2}{3};\ y_{3}=\frac {2}{3};\ y_{4}=-\frac {1}{2}\end{array}\right. $
（3）$\left\{\begin{array}{l} x_{1}=\sqrt {5},\ x_{2}=-\sqrt {5},\ x_{3}=2\sqrt {2},\ x_{4}=-2\sqrt {2},\\ y_{1}=\sqrt {5};\ y_{2}=-\sqrt {5};\ y_{3}=\sqrt {2};\ y_{4}=-\sqrt {2}\end{array}\right. $
（4）$\left\{\begin{array}{l} x_{1}=\frac {7}{3},\ x_{2}=\frac {13}{3},\\ y_{1}=-1;\ y_{2}=-4\end{array}\right. $
（5）$\left\{\begin{array}{l} x_{1}=0,\ x_{2}=\frac {1}{2},\ x_{3}=\frac {1}{2},\ x_{4}=1,\\ y_{1}=0;\ y_{2}=\frac {1}{2};\ y_{3}=-\frac {1}{2};\ y_{4}=0\end{array}\right. $
（6）$\left\{\begin{array}{l} x_{1}=\frac {\sqrt {6}}{2},\ x_{2}=\frac {\sqrt {6}}{2},\ x_{3}=-\frac {\sqrt {6}}{2},\ x_{4}=-\frac {\sqrt {6}}{2},\\ y_{1}=\frac {\sqrt {6}}{2};\ y_{2}=-\frac {\sqrt {6}}{2};\ y_{3}=\frac {\sqrt {6}}{2};\ y_{4}=-\frac {\sqrt {6}}{2}\end{array}\right. $
（7）$\left\{\begin{array}{l} x_{1}=-5,\ x_{2}=1,\\ y_{1}=4;\ y_{2}=1\end{array}\right. $
（8）$\left\{\begin{array}{l} x_{1}=-\frac {7}{4},\ x_{2}=5,\\ y_{1}=-\frac {3}{2};\ y_{2}=3\end{array}\right. $