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8.计算:(1)$(2a^{2})^{3}\cdot (-\frac {1}{2}a)^{2}= $____; (2)$(-2x^{3}y^{2})^{2}\cdot (-\frac {3}{2}x^{2}y^{3})^{2}= $____.
答案
8. (1)$2a^{8}$ (2)$9x^{10}y^{10}$
9.已知单项式$3x^{2}y^{3}与-2xy^{2}的积为mx^{3}y^{n}$,那么$m-n$的值为()
A.-11
B.5
C.1
D.-1
A.-11
B.5
C.1
D.-1
答案
9. A
10.(2025内江)若()$×(-xy)^{2}= 3x^{3}y^{3}$,则括号里应填的单项式是()
A.-3y
B.3xy
C.-3xy
D.$3x^{2}y$
A.-3y
B.3xy
C.-3xy
D.$3x^{2}y$
答案
10. B
11.(教材变式)计算:
(1)$3a^{2}\cdot 2a^{4}+(3a^{3})^{2}-14a^{6}$; (2)$(3xy^{2})^{2}+(4xy^{3})\cdot (-2xy)$;
(3)$(-3a^{2}b)^{3}-(-2a^{3}b)^{2}\cdot (-3b)$; (4)$(-2a)^{5}+(2a\cdot 3a)^{2}\cdot (-a)$.
(1)$3a^{2}\cdot 2a^{4}+(3a^{3})^{2}-14a^{6}$; (2)$(3xy^{2})^{2}+(4xy^{3})\cdot (-2xy)$;
(3)$(-3a^{2}b)^{3}-(-2a^{3}b)^{2}\cdot (-3b)$; (4)$(-2a)^{5}+(2a\cdot 3a)^{2}\cdot (-a)$.
答案
11. 解：(1)原式$=6a^{6}+9a^{6}-14a^{6}=a^{6}$；(2)原式$=9x^{2}y^{4}-8x^{2}y^{4}=x^{2}y^{4}$；(3)原式$=-27a^{6}b^{3}-4a^{6}b^{2}\cdot (-3b)=-27a^{6}b^{3}+12a^{6}b^{3}=-15a^{6}b^{3}$；(4)原式$=-32a^{5}-36a^{5}=-68a^{5}$。
12.先化简,再求值:$2x^{2}y\cdot (-2xy^{2})^{3}+(2xy)^{3}\cdot (-xy^{2})^{2}$,其中$x= 4,y= \frac {1}{4}$.
答案
12. 解：原式$=-16x^{5}y^{7}+8x^{5}y^{7}=-8x^{5}y^{7}$。当$x=4,y=\frac {1}{4}$时，原式$=-8×4^{5}×(\frac {1}{4})^{7}=-8×(\frac {1}{4})^{2}×(4×\frac {1}{4})^{5}=-8×\frac {1}{16}×1=-\frac {1}{2}$。
13.(教材变式)某中学一寝室前有一块长为$\frac {3}{2}x$,宽为$x$的空地,学校向全校师生征集这块地的绿化设计方案,并要求绿地面积不少于$x^{2}$,如图是学生小明的设计方案,阴影部分是绿地,试问小明的设计方案是否符合要求? 为什么?
答案
13. 解：$\frac {3}{2}x\cdot x-\frac {x}{2}\cdot \frac {3}{4}x-\frac {1}{2}π(\frac {x}{4})^{2}=\frac {3}{2}x^{2}-\frac {3}{8}x^{2}-\frac {π}{32}x^{2}=\frac {36-π}{32}x^{2}$。$\because π＜4$，$\therefore 36-π＞32$，$\therefore \frac {36-π}{32}x^{2}＞x^{2}$，$\therefore$ 小明的设计方案符合要求。
