《2025年超越训练八年级数学上册人教版四川专版》

12. 先化简，再求值：$-\frac {1}{2}x^{2}(2xy^{2}-4x^{2}y^{2})-4x^{2}y\cdot (-xy)$，其中$x$，$y$满足$(x-2)^{2}+|y+1|=0$.

答案： 12.解：原式$=3x^3y^2 + 2x^4y^2. \because (x - 2)^2 + $|y + 1|$ = 0, \therefore x = 2, y = -1, \therefore$原式=56.

13. 已知$(a^{m+1}b^{n+2})(a^{2n-1}b^{2m})=a^{5}b^{3}$，求$m^{n}$的值.

答案： 13.解：$\because (a^{m + 1}b^{n + 2})(a^{2n - 1}b^{2m}) = a^{m + 2n} \cdot b^{2m + n + 2} = a^5b^3, \therefore \begin{cases}m + 2n = 5, \\2m + n + 2 = 3,\end{cases}$解得$\begin{cases}m = -1, \\n = 3,\end{cases} \therefore m^n = (-1)^3 = -1.$

14. 两个完全相同的长方形按如图所示的方式放置，每个长方形的面积均为 28，图中阴影部分的面积为 20，求长方形的周长.

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答案： 14.解：设长方形的长为a，宽为b，则$ab = 28. \because$阴影部分的面积为$20, \therefore 28 × 2 - \frac{1}{2}(a + b)b - \frac{1}{2}ab = 20, \therefore 56 - \frac{1}{2}ab - \frac{1}{2}b^2 - \frac{1}{2}ab = 20, \therefore 56 - ab - \frac{1}{2}b^2 = 20, \therefore 56 - 28 - \frac{1}{2}b^2 = 20, \therefore b^2 = 16,$解得b = 4或b = -4(舍去)，$\therefore a = 28 ÷ 4 = 7, \therefore (4 + 7) × 2 = 22,$即长方形的周长为22.

15. 已知$-2x^{3m+1}y^{2n}$与$4x^{n-2}y^{6-m}$的积与$-4x^{4}y^{2}$是同类项.
(1)求$m$，$n$的值；
(2)先化简，再求值：$5m^{3}n\cdot (-3n)^{2}+(-6mn)^{2}\cdot (-mn)-mn^{3}\cdot (-4m)^{2}$.

答案： 15.解：$(1) -2x^{3m + 1}y^{2n} \cdot 4x^{n - 2}y^{6 - m} = -8x^{3m + n - 1}y^{2n - m + 6}.$
$\because -8x^{3m + n - 1}y^{2n - m + 6}$与$ -4x^4y^2$是同类项，$\therefore \begin{cases}3m + n - 1 = 4, \\2n - m + 6 = 2,\end{cases}$
解得$\begin{cases}m = 2, \\n = -1.\end{cases}(2)$原式$=45m^3n^3 - 36m^3n^3 - 16m^3n^3 = -7m^3n^3.$当m = 2,n = -1时，原式=56.

16. 已知$A=-2x(x^{2}-ax+1)+x(x+6)-2025$.
(1)若多项式$A$不含有$x^{2}$项，求$a$的值；
(2)在(1)的条件下，若$x^{2}-x-1=0$，求$A$的值.

答案： 16.解：$(1)A = -2x(x^2 - ax + 1) + x(x + 6) - 2025 = -2x^3 + 2ax^2 - 2x + x^2 + 6x - 2025 = -2x^3 + (2a + 1)x^2 + 4x - 2025 \because A$不含有$x^2$项，$\therefore 2a + 1 = 0，$解得$a = - \frac{1}{2}.(2)$由
(1)可得$A = -2x^3 + 4x - 2025. \because x^2 - x - 1 = 0, \therefore x^2 = x + 1,x^2 - x = 1, \therefore A = -2x \cdot x^2 + 4x - 2025 = -2(x^2 - x) - 2025 = -2 × 1 - 2025 = -2027.$