《2025年学霸题中题八年级数学上册人教版》

11. (1)若单项式$-6x^{2}y^{m}$与$\frac{1}{3}x^{n - 1}y^{3}$的和是单项式,则这两个单项式的积是$-2x^{4}y^{6}$;
(2) 改编题 若$(-2x^{2m - 1})^{3},(y^{n - 4})^{2}$与$7x^{1 - n}y^{m - 1}$的积与$x^{5}y^{3}$是同类项,则$m + n$的值为7.

答案：
(1)$ -2x^{4}y^{6} $
(2)7 解析:$ (-2x^{2m - 1})^{3}×(y^{n - 4})^{2}×7x^{1 - n}y^{m - 1}=-8x^{6m - 3}×y^{2n - 8}×7x^{1 - n}y^{m - 1}=-56x^{6m - 2 - n}y^{2n + m - 9} $
$ \because -56x^{6m - 2 - n}\cdot y^{2n + m - 9} $与 $ x^{5}y^{3} $是同类项，$ \therefore \left\{\begin{array}{l} 6m - 2 - n = 5\\ 2n + m - 9 = 3\end{array}\right. $，解得$ \left\{\begin{array}{l} m = 2\\ n = 5\end{array}\right. $，$ \therefore m + n = 7 $.

12. 计算:
(1)$(-4ab^{3})(-\frac{1}{8}ab)-(\frac{1}{2}ab^{2})^{2}$;
(2)$[2(a - b)^{3}][-3(a - b)^{2}][-\frac{2}{3}(b - a)]$;
(3)$5a^{3}b\cdot (-3b)^{2}+(-ab)(-6ab)^{2}$;
(4)$3x^{3}y^{3}\cdot (-\frac{2}{3}x^{2}y^{2})+(-\frac{1}{3}x^{2}y)^{3}\cdot 9xy^{2}$.

答案：
(1)原式$ = (-4ab^{3})(-\frac {1}{8}ab)-\frac {1}{4}a^{2}b^{4}=\frac {1}{2}a^{2}b^{4}-\frac {1}{4}a^{2}b^{4}=\frac {1}{4}a^{2}b^{4} $.
(2)原式$ = -4(a - b)^{6} $.
(3)原式$ = 5a^{3}b\cdot 9b^{2}-ab\cdot 36a^{2}b^{2}=45a^{3}b^{3}-36a^{3}b^{3}=9a^{3}b^{3} $.
(4)原式$ = 3x^{3}y^{3}\cdot (-\frac {2}{3}x^{2}y^{2})+(-\frac {1}{27}x^{6}y^{3})\cdot 9xy^{2}=-2x^{5}y^{5}-\frac {1}{3}x^{7}y^{5} $.

13. 三角$\begin{array}{c}\triangle \\ \end{array}\begin{array}{c}a\\ b\quad c\end{array}$表示$3abc$,方框$\begin{array}{c}\square \\ \end{array}\begin{array}{c}x\quad w\\ y\quad z\end{array}$表示$-4x^{y}w^{z}$,求$\begin{array}{c}\triangle \\ \end{array}\begin{array}{c}m\\ n\quad 3\end{array}\times \begin{array}{c}\square \\ \end{array}\begin{array}{c}n\quad m\\ 2\quad 5\end{array}$的值.

答案：
$ = 9mn×(-4n^{2}m^{5})=-36m^{6}n^{3} $.

14. 如图为小李家住房的结构图(单位:m),小李打算把卧室和客厅铺上木地板,请你帮他算一算,他至少应买木地板 (12xy)
A. $12xy m^{2}$
B. $10xy m^{2}$
C. $8xy m^{2}$
D. $6xy m^{2}$

答案： 14.A 解析:由题图知，卧室面积为 $ 2x\cdot 2y = 4xy(m^{2}) $，客厅面积为 $ 2x\cdot 4y = 8xy(m^{2})$，$ \therefore $他至少应买木地板 $ 4xy + 8xy = 12xy(m^{2}) $.

15. (1)已知$x^{3m}=2,y^{2m}=3$,求$(x^{2m})^{3}+(y^{m})^{6}-(x^{2}y)^{3m}\cdot y^{m}$的值;-5
(2)若$2 + 3+\cdots +n = m - 1,a^{2m}=4,b^{3m}=27$,求$(ab^{n})\cdot (a^{2}b^{n - 1})\cdot \cdots \cdot (a^{n - 1}b^{2})\cdot (a^{n}b)$的值.±6

答案：
(1)$ \because x^{3m}=2,y^{2m}=3,\therefore $原式$ = (x^{3m})^{2}+(y^{2m})^{3}-x^{6m}y^{3m}y^{m}=(x^{3m})^{2}+(y^{2m})^{3}-(x^{3m}y^{2m})^{2}=2^{2}+3^{3}-(2×3)^{2}=-5 $.
(2)由题意可得 $ 1 + 2 + 3 + \cdots + n = m,\therefore (ab^{n})\cdot (a^{2}b^{n - 1})\cdot \cdots \cdot (a^{n - 1}b^{2})\cdot (a^{n}b)=a^{1 + 2 + \cdots + n}\cdot b^{n + (n - 1) + \cdots + 1}=a^{m}b^{m}.\because a^{2m}=4,b^{3m}=27,\therefore (a^{m})^{2}=2^{2},(b^{m})^{3}=3^{3},\therefore a^{m}=\pm 2,b^{m}=3,\therefore $原式$ =\pm 6 $.