9. 计算:
(1)$(2x+3y)(3x-2y)$;
(2)$(2a-5)(3a+2)-6(a+1)(a-2)$.

(1) $(2x + 3y)(3x - 2y)$
$\begin{aligned}&=2x \cdot 3x + 2x \cdot (-2y) + 3y \cdot 3x + 3y \cdot (-2y)\\&=6x^2 - 4xy + 9xy - 6y^2\\&=6x^2 + 5xy - 6y^2\end{aligned}$
(2) $(2a - 5)(3a + 2) - 6(a + 1)(a - 2)$
$\begin{aligned}&=(2a \cdot 3a + 2a \cdot 2 - 5 \cdot 3a - 5 \cdot 2) - 6[(a \cdot a + a \cdot (-2) + 1 \cdot a + 1 \cdot (-2))]\\&=(6a^2 + 4a - 15a - 10) - 6(a^2 - 2a + a - 2)\\&=(6a^2 - 11a - 10) - 6(a^2 - a - 2)\\&=6a^2 - 11a - 10 - 6a^2 + 6a + 12\\&=-5a + 2\end{aligned}$

10. (1)先化简,再求值:$(x-1)(2x+1)-2(x-5)(x+2)$,其中$x= -2$;
(2)已知$x^{2}-5x= 14$,求$(x-1)(2x-1)-(x+1)^{2}+1$的值.

(1) 解：
原式
= $(x-1)(2x+1) - 2(x-5)(x+2)$
= $2x^2 + x - 2x - 1 - 2(x^2 - 3x - 10)$
= $2x^2 - x - 1 - 2x^2 + 6x + 20$
= $5x + 19$
当 $x = -2$ 时，
原式
= $5(-2) + 19$
= $-10 + 19$
= $9$
(2) 解：
原式
= $(x-1)(2x-1) - (x+1)^2 + 1$
= $2x^2 - x - 2x + 1 - (x^2 + 2x + 1) + 1$
= $2x^2 - 3x + 1 - x^2 - 2x - 1 + 1$
= $x^2 - 5x + 1$
由于 $x^2 - 5x = 14$，
所以原式
= $14 + 1$
= $15$