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19. (8分)
(1)计算：$(25x^{3}y - 10x^{2}y^{2} + 15xy^{3}) ÷ (-5xy)$；
(2)已知$(-3x^{4}y^{3})^{3} ÷ (-\frac{3}{2}x^{n}y^{2}) = -mx^{8}y^{7}$，求$m$，$n$的值.
(1)计算：$(25x^{3}y - 10x^{2}y^{2} + 15xy^{3}) ÷ (-5xy)$；
(2)已知$(-3x^{4}y^{3})^{3} ÷ (-\frac{3}{2}x^{n}y^{2}) = -mx^{8}y^{7}$，求$m$，$n$的值.
答案
(1)
$\begin{aligned}&(25x^{3}y - 10x^{2}y^{2} + 15xy^{3}) ÷ (-5xy)\\=&\frac{25x^{3}y}{-5xy}-\frac{10x^{2}y^{2}}{-5xy}+\frac{15xy^{3}}{-5xy}\\=& - 5x^{2}+2xy - 3y^{2}\end{aligned}$
(2)
$\begin{aligned}&(-3x^{4}y^{3})^{3}÷(-\frac{3}{2}x^{n}y^{2})\\=&-27x^{12}y^{9}÷(-\frac{3}{2}x^{n}y^{2})\\=&[-27÷(-\frac{3}{2})]×(x^{12 - n}y^{9 - 2})\\=&18x^{12 - n}y^{7}\end{aligned}$
因为$(-3x^{4}y^{3})^{3}÷(-\frac{3}{2}x^{n}y^{2})=-mx^{8}y^{7}$，所以$\begin{cases}12 - n = 8,\\18=-m.\end{cases}$
解得$\begin{cases}m = - 18,\\n = 4.\end{cases}$
$\begin{aligned}&(25x^{3}y - 10x^{2}y^{2} + 15xy^{3}) ÷ (-5xy)\\=&\frac{25x^{3}y}{-5xy}-\frac{10x^{2}y^{2}}{-5xy}+\frac{15xy^{3}}{-5xy}\\=& - 5x^{2}+2xy - 3y^{2}\end{aligned}$
(2)
$\begin{aligned}&(-3x^{4}y^{3})^{3}÷(-\frac{3}{2}x^{n}y^{2})\\=&-27x^{12}y^{9}÷(-\frac{3}{2}x^{n}y^{2})\\=&[-27÷(-\frac{3}{2})]×(x^{12 - n}y^{9 - 2})\\=&18x^{12 - n}y^{7}\end{aligned}$
因为$(-3x^{4}y^{3})^{3}÷(-\frac{3}{2}x^{n}y^{2})=-mx^{8}y^{7}$，所以$\begin{cases}12 - n = 8,\\18=-m.\end{cases}$
解得$\begin{cases}m = - 18,\\n = 4.\end{cases}$
20. (8分)计算：
(1)$(\frac{3}{4}x^{2}y - \frac{1}{2}xy^{2} - \frac{5}{2}y^{3}) · (-4xy^{2})$；
(2)$(2a - 3b)(2a^{2} + 6ab + 5b^{2})$.
(1)$(\frac{3}{4}x^{2}y - \frac{1}{2}xy^{2} - \frac{5}{2}y^{3}) · (-4xy^{2})$；
(2)$(2a - 3b)(2a^{2} + 6ab + 5b^{2})$.
答案
(1)
$\begin{aligned}&(\frac{3}{4}x^{2}y - \frac{1}{2}xy^{2} - \frac{5}{2}y^{3}) · (-4xy^{2})\\=&\frac{3}{4}x^{2}y·(-4xy^{2}) - \frac{1}{2}xy^{2}·(-4xy^{2}) - \frac{5}{2}y^{3}·(-4xy^{2})\\=&-3x^{3}y^{3} + 2x^{2}y^{4} + 10xy^{5}\end{aligned}$
(2)
$\begin{aligned}&(2a - 3b)(2a^{2} + 6ab + 5b^{2})\\=&2a·2a^{2} + 2a·6ab + 2a·5b^{2} - 3b·2a^{2} - 3b·6ab - 3b·5b^{2}\\=&4a^{3} + 12a^{2}b + 10ab^{2} - 6a^{2}b - 18ab^{2} - 15b^{3}\\=&4a^{3} + (12a^{2}b - 6a^{2}b) + (10ab^{2} - 18ab^{2}) - 15b^{3}\\=&4a^{3} + 6a^{2}b - 8ab^{2} - 15b^{3}\end{aligned}$
$\begin{aligned}&(\frac{3}{4}x^{2}y - \frac{1}{2}xy^{2} - \frac{5}{2}y^{3}) · (-4xy^{2})\\=&\frac{3}{4}x^{2}y·(-4xy^{2}) - \frac{1}{2}xy^{2}·(-4xy^{2}) - \frac{5}{2}y^{3}·(-4xy^{2})\\=&-3x^{3}y^{3} + 2x^{2}y^{4} + 10xy^{5}\end{aligned}$
(2)
$\begin{aligned}&(2a - 3b)(2a^{2} + 6ab + 5b^{2})\\=&2a·2a^{2} + 2a·6ab + 2a·5b^{2} - 3b·2a^{2} - 3b·6ab - 3b·5b^{2}\\=&4a^{3} + 12a^{2}b + 10ab^{2} - 6a^{2}b - 18ab^{2} - 15b^{3}\\=&4a^{3} + (12a^{2}b - 6a^{2}b) + (10ab^{2} - 18ab^{2}) - 15b^{3}\\=&4a^{3} + 6a^{2}b - 8ab^{2} - 15b^{3}\end{aligned}$
