15. (2024·内江中考)已知实数a、b满足$ab = 1$,那么$\frac{1}{a^{2} + 1}+\frac{1}{b^{2} + 1}$的值为________.

1

16. (2023·日照中考改编)若关于x的方程$\frac{x}{x - 1}-2=\frac{3m}{2x - 2}$的解为正数,则m的取值范围是________.

$m ＜ \frac{4}{3}$且$m \neq \frac{2}{3}$

17. 现有甲、乙、丙三种糖混合而成的什锦糖50千克,其中各种糖的质量和单价如下表.
　　
商店以糖的平均价(平均价 = 混合糖的总价格÷混合糖的总质量)作为什锦糖的单价,要使什锦糖的单价每千克提高1元,则需再加入丙种糖________千克.

12.5 解析：设再加入丙种糖$x$千克，则什锦糖的总质量为$(20 + 10 + 20 + x)$千克，根据题意得$\frac{20\times15 + 10\times20 + 20\times25}{20 + 10 + 20} + 1 = \frac{20\times15 + 10\times20 + (20 + x)\times25}{20 + 10 + 20 + x}$，解得$x = 12.5$，经检验$x = 12.5$是原分式方程的解，$\therefore$需再加入丙种糖 12.5 千克.

18. 已知$a + b = m$,$a - b = n$,则$\frac{na + m^{2}}{b}-\frac{nb + m^{2}}{a}$的值为________(用含m、n的式子表示).
　　　　　　　　　　　　　　

$\frac{4mn}{m - n}$ 解析：$\frac{na + m^2}{b} - \frac{nb + m^2}{a} = \frac{na^2 + am^2}{ab} - \frac{nb^2 + bm^2}{ab} = \frac{na^2 + am^2 - (nb^2 + bm^2)}{ab} = \frac{na^2 + am^2 - nb^2 - bm^2}{ab} = \frac{n(a^2 - b^2) + m^2(a - b)}{ab} = \frac{n(a - b)(a + b) + m^2(a - b)}{ab}$.$\because a + b = m$，$a - b = n$，故$m + n = a + b + a - b = 2a$，$m - n = a + b - (a - b) = 2b$，$\therefore a = \frac{m + n}{2}$，$b = \frac{m - n}{2}$，$ab = \frac{(m + n)(m - n)}{4}$，将$a + b = m$，$a - b = n$，$ab = \frac{(m + n)(m - n)}{4}$代入原式，原式$= \frac{mn^2 + nm^2}{\frac{(m + n)(m - n)}{4}} = \frac{mn(n + m)}{\frac{(m + n)(m - n)}{4}} = \frac{4mn(n + m)}{(m + n)(m - n)} = \frac{4mn}{m - n}$.

三、解答题(共46分)

19. (8分)计算:
(1) (镇江中考)$(x^{2} - 1)\div(1-\frac{1}{x})-x$;
(2) (2023·泸州中考)$(\frac{4m + 5}{m + 1}+m - 1)\div\frac{m + 2}{m + 1}$.

(1)原式$= (x + 1)(x - 1) \cdot \frac{x}{x - 1} - x = x(x + 1) - x = x^2 + x - x = x^2$.
(2)原式$= (\frac{4m + 5}{m + 1} + \frac{m^2 - 1}{m + 1}) \cdot \frac{m + 1}{m + 2} = \frac{(m + 2)^2}{m + 1} \cdot \frac{m + 1}{m + 2} = m + 2$.

20. (10分)解方程:
(1) (2024·广州中考)$\frac{1}{2x - 5}=\frac{3}{x}$;
(2) $\frac{x}{x - 1}-\frac{3}{(x - 1)(x + 2)}=1$.

(1)方程两边同乘$x(2x - 5)$，得$x = 3(2x - 5)$，解得$x = 3$，检验：当$x = 3$时，$x(2x - 5) \neq 0$，$\therefore$该分式方程的解为$x = 3$.
(2)方程两边同乘$(x - 1)(x + 2)$，得$x^2 + 2x - 3 = (x - 1)(x + 2)$，解得$x = 1$，检验：当$x = 1$时，$(x - 1)(x + 2) = 0$，$\therefore x = 1$是原方程的增根，即原方程无解.

21. (8分) (2024·广安中考)先化简$(a + 1-\frac{3}{a - 1})\div\frac{a^{2} + 4a + 4}{a - 1}$,再从-2,0,1,2中选取一个适合的数代入求值.

原式$= (\frac{a^2 - 1}{a - 1} - \frac{3}{a - 1}) \div \frac{(a + 2)^2}{a - 1} = \frac{(a + 2)(a - 2)}{a - 1} \cdot \frac{a - 1}{(a + 2)^2} = \frac{a - 2}{a + 2}$，$a \neq 1$且$a \neq -2$.
①当$a = 0$时，原式$= -1$；
②当$a = 2$时，原式$= 0$.