12. 通分：
(1)$\frac {1}{3x^{2}},\frac {5}{12xy}$；
(2)$\frac {1}{x-1},\frac {1}{x^{2}-1},\frac {1}{x^{2}+x}$；
(3)$\frac {x}{x-y},\frac {y}{x^{2}+2xy+y^{2}},\frac {2}{y^{2}-x^{2}}$；
(4)$\frac {x}{x^{2}-1},\frac {2}{x^{2}-2x+1}$.
答案：解:
(1)$\frac{1}{3x^{2}}=\frac{4y}{12x^{2}y}$,$\frac{5}{12xy}=\frac{5x}{12x^{2}y}$.
(2)$\frac{1}{x-1}=\frac{x(x+1)}{x(x+1)(x-1)}$,$\frac{1}{x^{2}-1}=\frac{x}{x(x+1)(x-1)}$,$\frac{1}{x^{2}+x}=\frac{x-1}{x(x+1)(x-1)}$.
(3)$\frac{x}{x-y}=\frac{x(x+y)^{2}}{(x-y)(x+y)^{2}}$,$\frac{y}{x^{2}+2xy+y^{2}}=\frac{y(x-y)}{(x-y)(x+y)^{2}}$,$\frac{2}{y^{2}-x^{2}}=-\frac{2(x+y)}{(x-y)(x+y)^{2}}$.
(4)$\frac{x}{x^{2}-1}=\frac{x(x-1)}{(x-1)^{2}(x+1)}$,$\frac{2}{x^{2}-2x+1}=\frac{2(x+1)}{(x-1)^{2}(x+1)}$.
13. 先化简,再求值：$\frac {3x^{2}-xy}{9x^{2}-6xy+y^{2}}$,其中$x= \frac {4}{3},y= -\frac {2}{3}$.
答案：解:原式=$\frac{x}{3x-y}$.当$x=\frac{4}{3}$,$y=-\frac{2}{3}$时,原式=$\frac{2}{7}$.
解析：
解:原式=$\frac{x(3x-y)}{(3x-y)^2}=\frac{x}{3x-y}$.当$x=\frac{4}{3}$,$y=-\frac{2}{3}$时,原式=$\frac{\frac{4}{3}}{3×\frac{4}{3}-(-\frac{2}{3})}=\frac{\frac{4}{3}}{\frac{14}{3}}=\frac{4}{14}=\frac{2}{7}$.
14. (2024 春·雅安期末)我们知道,分数可分为真分数和假分数,假分数都可化为带分数. 如$\frac {8}{3}= \frac {6+2}{3}= 2+\frac {2}{3}= 2\frac {2}{3}$. 我们定义：在分式中,对于只含有一个字母的分式,当分子的次数大于或等于分母的次数时,我们称之为“假分式”；当分子的次数小于分母的次数时,我们称之为“真分式”. 如$\frac {x-1}{x+1},\frac {x^{2}+2x-2}{x+1}$这样的分式是假分式；$\frac {3}{x+1},\frac {2x}{x^{2}+1}$这样的分式是真分式. 类似地,假分式也可以化为带分式. 如$\frac {x-1}{x+1}= \frac {x+1-2}{x+1}= 1-\frac {2}{x+1},\frac {x^{2}+3}{x+1}= \frac {x^{2}-1+4}{x+1}= \frac {(x+1)(x-1)+4}{x+1}= x-1+\frac {4}{x+1}$.
解决下列问题：
(1) 分式$\frac {1}{2x}$是______(填“真”或“假”)分式；
(2) 将假分式$\frac {x^{2}-1}{x+2}$化为带分式；
(3) 若$x$为整数,分式$\frac {2x^{2}-3x-1}{x+2}$的值为整数,求所有符合条件的$x$的值.
真
解:$\frac{x^{2}-1}{x+2}=\frac{x^{2}-4+3}{x+2}=\frac{(x+2)(x-2)+3}{x+2}=x-2+\frac{3}{x+2}$.
解:原式=$2x-7+\frac{13}{x+2}$.$\because$分式的值为整数,$\therefore x+2=\pm1$或$\pm13$,$\therefore x=-1$或$-3$或11或$-15$.
答案：
(1)真
(2)解:$\frac{x^{2}-1}{x+2}=\frac{x^{2}-4+3}{x+2}=\frac{(x+2)(x-2)+3}{x+2}=x-2+\frac{3}{x+2}$.
(3)解:原式=$2x-7+\frac{13}{x+2}$.$\because$分式的值为整数,$\therefore x+2=\pm1$或$\pm13$,$\therefore x=-1$或$-3$或11或$-15$.
