20. (2025·宿豫期中)先化简，再求值：$\frac{3x + 9}{x} ÷ (x - \frac{9}{x})$，其中$x = \frac{5}{2}$.
答案：20. 解：原式$ = \frac{3(x + 3)}{x} ÷ \frac{x^{2} - 9}{x} = \frac{3(x + 3)}{x} · \frac{x}{(x + 3)(x - 3)} = \frac{3}{x - 3} $
当$ x = \frac{5}{2} $时，原式$ = \frac{3}{\frac{5}{2} - 3} = -6 $
当$ x = \frac{5}{2} $时，原式$ = \frac{3}{\frac{5}{2} - 3} = -6 $
21. 解方程：(1)(2025·广州)$\frac{1}{2x - 5} = \frac{3}{x}$；
(2)(2025·南通)$\frac{x}{x + 1} - 1 = \frac{2x}{3x + 3}$.
(2)(2025·南通)$\frac{x}{x + 1} - 1 = \frac{2x}{3x + 3}$.
答案：21. 解：(1)去分母，得$ x = 3(2x - 5) $，去括号，得$ x = 6x - 15 $
移项、合并同类项，得$ -5x = -15 $
系数化为1，得$ x = 3 $
检验：当$ x = 3 $时，$ x(2x - 5) ≠ 0 $
故原方程的解为$ x = 3 $
(2)去分母，得$ 3x - 3(x + 1) = 2x $
去括号，得$ 3x - 3x - 3 = 2x $，解得$ x = -\frac{3}{2} $
检验：当$ x = -\frac{3}{2} $时，$ 3(x + 1) ≠ 0 $
∴分式方程的解为$ x = -\frac{3}{2} $
移项、合并同类项，得$ -5x = -15 $
系数化为1，得$ x = 3 $
检验：当$ x = 3 $时，$ x(2x - 5) ≠ 0 $
故原方程的解为$ x = 3 $
(2)去分母，得$ 3x - 3(x + 1) = 2x $
去括号，得$ 3x - 3x - 3 = 2x $，解得$ x = -\frac{3}{2} $
检验：当$ x = -\frac{3}{2} $时，$ 3(x + 1) ≠ 0 $
∴分式方程的解为$ x = -\frac{3}{2} $
22. 已知$M = \frac{x + 1}{2}$，$N = \frac{2x}{x + 1}$.
(1)当$x ＞ 0$时，判断$M$与$N$的大小关系，并说明理由；
(2)设$y = \frac{2}{M} + N$.若$x$是整数，求$y$的整数值.
(1)当$x ＞ 0$时，判断$M$与$N$的大小关系，并说明理由；
(2)设$y = \frac{2}{M} + N$.若$x$是整数，求$y$的整数值.
答案：22. 解：(1)$ M ≥ N $，理由：
$ \because M - N = \frac{x + 1}{2} - \frac{2x}{x + 1} = \frac{(x + 1)^{2} - 4x}{2(x + 1)} = \frac{(x - 1)^{2}}{2(x + 1)} $
又$ \because x ＞ 0 $，$ \therefore x + 1 ＞ 0 $，$ (x - 1)^{2} ≥ 0 $
$ \therefore \frac{(x - 1)^{2}}{2(x + 1)} ≥ 0 $，$ \therefore M - N ≥ 0 $，$ \therefore M ≥ N $
(2)$ y = \frac{2}{M} + N = \frac{2}{\frac{x + 1}{2}} + \frac{2x}{x + 1} = \frac{4}{x + 1} + \frac{2x}{x + 1} = \frac{2(x + 1) + 2}{x + 1} = 2 + \frac{2}{x + 1} $
$ \because x $，$ y $均为整数，$ \therefore x + 1 $是2的因数
$ \therefore x + 1 = \pm 1 $，$ \pm 2 $
对应的$ y $的值为$ y = 2 + 2 = 4 $或$ y = 2 + (-2) = 0 $或$ y = 2 + 1 = 3 $或$ y = 2 - 1 = 1 $
$ \therefore y $的整数值为4，0，3，1
$ \because M - N = \frac{x + 1}{2} - \frac{2x}{x + 1} = \frac{(x + 1)^{2} - 4x}{2(x + 1)} = \frac{(x - 1)^{2}}{2(x + 1)} $
又$ \because x ＞ 0 $，$ \therefore x + 1 ＞ 0 $，$ (x - 1)^{2} ≥ 0 $
$ \therefore \frac{(x - 1)^{2}}{2(x + 1)} ≥ 0 $，$ \therefore M - N ≥ 0 $，$ \therefore M ≥ N $
(2)$ y = \frac{2}{M} + N = \frac{2}{\frac{x + 1}{2}} + \frac{2x}{x + 1} = \frac{4}{x + 1} + \frac{2x}{x + 1} = \frac{2(x + 1) + 2}{x + 1} = 2 + \frac{2}{x + 1} $
$ \because x $，$ y $均为整数，$ \therefore x + 1 $是2的因数
$ \therefore x + 1 = \pm 1 $，$ \pm 2 $
对应的$ y $的值为$ y = 2 + 2 = 4 $或$ y = 2 + (-2) = 0 $或$ y = 2 + 1 = 3 $或$ y = 2 - 1 = 1 $
$ \therefore y $的整数值为4，0，3，1