《2025年基础训练大象出版社八年级数学上册人教版》

22. (10 分)对于$m + n,\frac{1}{m} + \frac{1}{n},m^2 + n^2$等代数式,交换m和n的位置,式子的值不变,我们把这样的式子叫作完美对称式.若关于x,y的分式$\frac{y}{x} - \frac{mx}{y}$是完美对称式,则：(1)m的值为-1；(2) 若完美对称式$\frac{y}{x} - \frac{mx}{y}$满足$\frac{y}{x} - \frac{mx}{y} = 5,$且xy = 3,x ＞ y,求x - y的值.$\because $完美对称式$\frac{y}{x}-\frac{mx}{y}$满足$\frac{y}{x}-\frac{mx}{y}=5,\therefore m=-1.\therefore \frac{y}{x}+\frac{x}{y}=5.\therefore \frac{x^2+y^2}{xy}=5.\therefore x^2+y^2=5xy.$又$\because xy=3,\therefore x^2+y^2=15.\therefore (x-y)^2=x^2+y^2-2xy=15-2×3=9.\because x＞y,$∴$ x-y＞0.\therefore x-y=3,$即为所求.

答案： （1）-1（2）$\because$ 完美对称式$\frac{y}{x}-\frac{mx}{y}$满足$\frac{y}{x}-\frac{mx}{y}=5$,$\therefore m=-1$.$\therefore \frac{y}{x}+\frac{x}{y}=5$.$\therefore \frac{x^2+y^2}{xy}=5$.$\therefore x^2+y^2=5xy$.又$\because xy=3$,$\therefore x^2+y^2=15$.$\therefore (x-y)^2=x^2+y^2-2xy=15-2×3=9$.$\because x＞y$,$∴ x-y＞0$.$\therefore x-y=3$,即为所求.