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第79页

信息发布者：

 解：

 $\begin{aligned}&(36x^{4}y^{3}-24x^{3}y^{2}+3x^{2}y^{2})\div(-6x^{2}y^{2})\\=&36x^{4}y^{3}\div(-6x^{2}y^{2})-24x^{3}y^{2}\div(-6x^{2}y^{2})+3x^{2}y^{2}\div(-6x^{2}y^{2})\\=& - 6x^{2}y + 4x-\frac{1}{2}\end{aligned}$

 解：

 $\begin{aligned}&(x + 1)(x - 2)+(2x + 1)(x - 5)-3(x^{2}-6x - 1)\\=&x^{2}-2x+x - 2+(2x^{2}-10x+x - 5)-3x^{2}+18x + 3\\=&x^{2}-x - 2+2x^{2}-9x - 5-3x^{2}+18x + 3\\=&(x^{2}+2x^{2}-3x^{2})+(-x-9x + 18x)+(-2-5 + 3)\\=&8x-4\end{aligned}$

 当$x =-\frac{1}{2}$时，原式$=8\times(-\frac{1}{2})-4=-4 - 4=-8$

 解：

 原不等式可化为$2x-x^{2}+5x - x + 5＞x - x^{2}+3$

 移项可得：$2x-x^{2}+5x - x - x + x^{2}＞3 - 5$

 合并同类项得：$5x＞-2$

 解得$x＞-\frac{2}{5}$

 (1)解：

 $\begin{aligned}\because(x + a)(x + 6)&=x^{2}+6x+ax + 6a\\&=x^{2}+(6 + a)x+6a\end{aligned}$

 又$\because x^{2}+(6 + a)x+6a=x^{2}+8x + 12$

 $\therefore\begin{cases}6 + a = 8\\6a = 12\end{cases}$

 由$6 + a = 8$得$a = 2，$将$a = 2$代入$6a = 12，$等式成立，所以$a = 2。$

 (2)当$a = 2，$$b=-3$时，

 $\begin{aligned}&(x + a)(x + b)\\=&(x + 2)(x - 3)\\=&x^{2}-3x+2x - 6\\=&x^{2}-x - 6\end{aligned}$

$x^{3}+1$

$m^{3}+8$

$8a^{3}+1$

解​$：(2)$​规律：​$(a+b)(a^2-ab+b^2)=a^3+b^3.$​
证明：​$(a+b)(a^2-ab+b^2)$​
​$=a^3-a^2b+ab^2+a^2b-ab^2+b^3$​
​$=a^3+b^3.$​
​$(3)∵x+y=2，$​​$xy=-3，$​
​$∴x^2+y^2=(x+y)^2-2xy=10，$​
​$∴x^3+y^3$​
​$=(x+y)(x^2-xy+y^2)$​
​$=26.$​