13. 已知关于$x$的一元二次方程$(a - 1)x^{a(3 - a)} + (2a + 1)x + 5 = 0$.
（1）求$a$的值；
（2）用配方法解这个方程.

(1) $\because$方程$(a - 1)x^{a(3 - a)} + (2a + 1)x + 5 = 0$是关于$x$的一元二次方程,$\therefore a(3 - a)=2$.解得$a=1$或$a=2$. $\because a - 1 ≠ 0, \therefore a ≠ 1$. $\therefore a$的值为2
(2) 当$a=2$时,方程为$x^2 + 5x + 5 = 0$,即$x^2 + 5x =-5$.配方,得$x^2 + 5x + \frac{25}{4}=-5 + \frac{25}{4}$,即$(x + \frac{5}{2})^2=\frac{5}{4}$.直接开平方,得$x + \frac{5}{2}=\pm \frac{\sqrt{5}}{2}. \therefore x_1=\frac{-5 + \sqrt{5}}{2}, x_2=\frac{-5 - \sqrt{5}}{2}$

14. 阅读材料：
若$m^2 - 2mn + 2n^2 - 8n + 16 = 0$，求$m$、$n$的值.
解：$\because m^2 - 2mn + 2n^2 - 8n + 16 = 0$，
$\therefore (m^2 - 2mn + n^2) + (n^2 - 8n + 16) = 0$.
$\therefore (m - n)^2 + (n - 4)^2 = 0$.
$\therefore (m - n)^2 = 0$，$(n - 4)^2 = 0$.
$\therefore m = 4$，$n = 4$.
根据你的观察，探究下面的问题：
（1）若$a^2 + b^2 - 2a + 1 = 0$，则$a =$，$b =$.
（2）已知$x^2 + 2y^2 + 2xy - 6y + 9 = 0$，求$x^y$的值.
（3）已知$△ ABC$的三边长$a$、$b$、$c$都是正整数，且满足$2a^2 + b^2 - 4a - 6b + 11 = 0$，求$△ ABC$的周长.

14. (1) 1 0
(2) $\because x^2 + 2y^2 + 2xy - 6y + 9 = 0, \therefore (x^2 + y^2 + 2xy) + (y^2 - 6y + 9) = 0$,即$(x + y)^2 + (y - 3)^2 = 0. \therefore x + y = 0, y - 3 = 0. \therefore x = -3, y = 3. \therefore x^y = (-3)^3 = -27$
(3) $\because 2a^2 + b^2 - 4a - 6b + 11 = 0, \therefore (2a^2 - 4a + 2) + (b^2 - 6b + 9) = 0. \therefore 2(a - 1)^2 + (b - 3)^2 = 0. \therefore a - 1 = 0, b - 3 = 0$.解得$a=1, b=3$. 由三角形的三边关系可知,$△ ABC$的三边长分别为1、3、3. $\therefore △ ABC$的周长为$1 + 3 + 3 =7$