11.(7分)已知$n$为正整数,且$x^{2n}=7$,求$(3x^{3n})^{2}-4(x^{2})^{2n}$的值.

$2891$

$(3x^{3n})^{2}-4(x^{2})^{2n}$
$=9x^{6n}-4x^{4n}$
$=9(x^{2n})^{3}-4(x^{2n})^{2}$
因为$x^{2n}=7$，
所以原式$=9×7^{3}-4×7^{2}$
$=9×343 - 4×49$
$=3087 - 196$
$=2891$

12.(7分)若关于$x$的多项式$(x^{2}+x - n)(mx - 3)$的展开式中不含$x^{2}$项和常数项，求$m,n$的值.

$(x^{2}+x - n)(mx - 3)$
$=x^{2}(mx - 3)+x(mx - 3)-n(mx - 3)$
$=mx^{3}-3x^{2}+mx^{2}-3x - mnx + 3n$
$=mx^{3}+(m - 3)x^{2}+(-3 - mn)x + 3n$
因为展开式中不含$x^{2}$项和常数项，所以：
$\begin{cases}m - 3 = 0 \\ 3n = 0\end{cases}$
解得$\begin{cases}m = 3 \\ n = 0\end{cases}$
结论：$m=3$，$n=0$

13.(8分)实数$x,y$满足条件$|x + 2|+(y + 1)^{2}=0$,求$(-2xy)^{2} · (-y^{2}) · 6xy^{2}$的值.

因为|x + 2|+(y + 1)²=0，且|x + 2|≥0，(y + 1)²≥0，所以x + 2=0，y + 1=0，解得x=-2，y=-1。
原式=(-2xy)²·(-y²)·6xy²
=4x²y²·(-y²)·6xy²
=4×(-1)×6·x²·x·y²·y²·y²
=-24x³y⁶
当x=-2，y=-1时，
原式=-24×(-2)³×(-1)⁶
=-24×(-8)×1
=192

14.(8分)(1)化简求值:$x(x^{2}-4)-(x + 3)(x^{2}-3x + 2)-2x(x - 2)$,其中$x = \frac{3}{2}$.
(2)已知$x^{2}-2x = 1$,求$(x - 1)(3x + 1)-(x + 1)^{2}$的值.

(1) $0$；(2) $0$

(1) $x(x^{2}-4)-(x + 3)(x^{2}-3x + 2)-2x(x - 2)$
$\begin{aligned}&=x^3 - 4x -[(x(x^2 - 3x + 2) + 3(x^2 - 3x + 2))]-2x^2 + 4x\\&=x^3 - 4x -[(x^3 - 3x^2 + 2x + 3x^2 - 9x + 6)] - 2x^2 + 4x\\&=x^3 - 4x - (x^3 - 7x + 6) - 2x^2 + 4x\\&=x^3 - 4x - x^3 + 7x - 6 - 2x^2 + 4x\\&=-2x^2 + 7x - 6\end{aligned}$
当$x = \frac{3}{2}$时，
$\begin{aligned}&-2×(\frac{3}{2})^2 + 7×\frac{3}{2} - 6\\&=-2×\frac{9}{4} + \frac{21}{2} - 6\\&=-\frac{9}{2} + \frac{21}{2} - 6\\&=6 - 6\\&=0\end{aligned}$
(2) $(x - 1)(3x + 1)-(x + 1)^{2}$
$\begin{aligned}&=3x^2 + x - 3x - 1 - (x^2 + 2x + 1)\\&=3x^2 - 2x - 1 - x^2 - 2x - 1\\&=2x^2 - 4x - 2\\&=2(x^2 - 2x) - 2\end{aligned}$
因为$x^2 - 2x = 1$，所以原式$=2×1 - 2 = 0$