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第75页

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$-\frac{3}{2}$

 解：$\begin{aligned}&(x - 2y)(x^{2}+2xy + 4y^{2})-(x + 2y)(x^{2}-4y^{2})\\=&x^{3}-(2y)^{3}-(x^{3}-4xy^{2}+2x^{2}y-8y^{3})\\=&x^{3}-8y^{3}-x^{3}+4xy^{2}-2x^{2}y + 8y^{3}\\=&4xy^{2}-2x^{2}y\end{aligned}$

 当$x = 4，$$y=\frac{1}{2}$时，

 $\begin{aligned}&4\times4\times(\frac{1}{2})^{2}-2\times4^{2}\times\frac{1}{2}\\=&4\times4\times\frac{1}{4}-2\times16\times\frac{1}{2}\\=&4 - 16\\=& - 12\end{aligned}$

 解：

 (1)由题意，得$(2x - a)(3x + b)=6x^{2}+2bx-3ax - ab=6x^{2}+(2b - 3a)x - ab = 6x^{2}+11x - 10，$$(2x + a)(x + b)=2x^{2}+2bx+ax + ab=2x^{2}+(2b + a)x + ab = 2x^{2}-9x + 10.$

 所以$\begin{cases}2b - 3a = 11\\2b + a = - 9\end{cases}，$

 用$2b - 3a = 11$减去$2b + a = - 9$得：

 $\begin{aligned}(2b - 3a)-(2b + a)&=11-(-9)\\2b - 3a - 2b - a&=11 + 9\\-4a&=20\\a&=-5\end{aligned}$

 把$a = - 5$代入$2b + a = - 9$得：$2b-5=-9，$$2b=-4，$$b = - 2.$

 (2)$(2x - 5)(3x - 2)=6x^{2}-4x-15x + 10=6x^{2}-19x + 10.$

 解：根据题意，得$(x^{3}+mx + n)(x^{2}-3x + 4)=x^{5}-3x^{4}+4x^{3}+mx^{3}-3mx^{2}+4mx+nx^{2}-3nx + 4n=x^{5}-3x^{4}+(4 + m)x^{3}+(n - 3m)x^{2}+(4m - 3n)x + 4n.$

 因为展开的结果中不含$x^{3}$和$x^{2}$的项，所以$\begin{cases}4 + m = 0\\n - 3m = 0\end{cases}，$

 由$4 + m = 0$得$m = - 4，$

 把$m = - 4$代入$n - 3m = 0$得$n-3\times(-4)=0，$$n + 12 = 0，$$n = - 12.$