13. 先化简，再求值：$(x + 1)(x - 1)-(x + 3)^{2}+2x^{2}$，其中$x^{2}-3x - 2 = 0$.

13. 原式$= x^2 - 1 - x^2 - 6x - 9 + 2x^2 = 2x^2 - 6x - 10$，$\because x^2 - 3x - 2 = 0$，$\therefore x^2 - 3x = 2$，$\therefore$原式$= 2(x^2 - 3x) - 10 = 2×2 - 10 = -6$.

14. 已知$x + y = 3$，$xy = -10$.
(1)求$(3 - x)(3 - y)$的值.
(2)求$x^{2}+3xy + y^{2}$的值.

14. (1)$\because x + y = 3$，$xy = -10$，$\therefore$原式$= 9 - 3y - 3x + xy = 9 - 3(x + y) + xy = 9 - 3×3 - 10 = 9 - 9 - 10 = -10$. (2)$\because x + y = 3$，$xy = -10$，$\therefore$原式$= (x + y)^2 + xy = 9 - 10 = -1$.

15. 两个边长分别为$a$和$b$的正方形如图放置（图1），其未叠合部分（阴影）面积为$S_{1}$；若再在图1中大正方形的右下角摆放一个边长为$b$的小正方形（如图2），两个小正方形叠合部分（阴影）面积为$S_{2}$.
(1)用含$a$、$b$的代数式分别表示$S_{1}$、$S_{2}$.
(2)若$a + b = 10$，$ab = 20$，求$S_{1}+S_{2}$的值.
(3)当$S_{1}+S_{2}=30$时，求图3中阴影部分的面积$S_{3}$.

15. (1)由图可得，$S_1 = a^2 - b^2$，$S_2 = a^2 - a(a - b) - b(a - b) - b(a - b) = 2b^2 - ab$. (2)$\because a + b = 10$，$ab = 20$，$\therefore S_1 + S_2 = a^2 - b^2 + 2b^2 - ab = a^2 + b^2 - ab = (a + b)^2 - 3ab = 100 - 3×20 = 40$. (3)由图可得，$S_3 = a^2 + b^2 - \frac{1}{2}b(a + b) - \frac{1}{2}a^2 = \frac{1}{2}(a^2 + b^2 - ab)$，$\because S_1 + S_2 = a^2 + b^2 - ab = 30$，$\therefore S_3 = \frac{1}{2}×30 = 15$.