搜索
三、计算题(本大题共8小题,共46分)
答案：
19. (4分)先化简,再求值:$(2x - 1)^{2} - (3x + 1)(3x - 1) + 5x(x - 1)$,其中$x = \frac{1}{2}$.
解:原式$ = 4x^{2} - 4x + 1 - 9x^{2} + 1 + 5x^{2} - 5x = 2 - 9x$,当$x = \frac{1}{2}$时,原式$ = -\frac{5}{2}$.
解:原式$ = 4x^{2} - 4x + 1 - 9x^{2} + 1 + 5x^{2} - 5x = 2 - 9x$,当$x = \frac{1}{2}$时,原式$ = -\frac{5}{2}$.
答案：
原式$=4x^{2}-4x + 1 - 9x^{2}+1 + 5x^{2}-5x = 2 - 9x$，
当 $x=\frac{1}{2}$ 时，原式$=-\frac{5}{2}$.
当 $x=\frac{1}{2}$ 时，原式$=-\frac{5}{2}$.
20. (4分)(2024·山东济宁微山期末)已知关于$x$的代数式$(x + 2m) \cdot (x^{2} - x + \frac{1}{2}n)$中不含$x$项与$x^{2}$项.
(1)求$m$,$n$的值;
(2)求代数式$m^{2023}n^{2024}$的值.
解:(1)$(x + 2m)(x^{2} - x + \frac{1}{2}n)$
$ = x^{3} - x^{2} + \frac{1}{2}nx + 2mx^{2} - 2mx + mn$
$ = x^{3} + (2m - 1)x^{2} + (\frac{1}{2}n - 2m)x + mn$.
$\because$不含$x$项与$x^{2}$项,
$\therefore\begin{cases}2m - 1 = 0\\\frac{1}{2}n - 2m = 0\end{cases}$,解得$\begin{cases}m = \frac{1}{2}\\n = 2\end{cases}$.
(2)$m^{2023}n^{2024} = (\frac{1}{2})^{2023} \cdot 2^{2024} = (\frac{1}{2} \times 2)^{2023} \times 2 = 2$.
(1)求$m$,$n$的值;
(2)求代数式$m^{2023}n^{2024}$的值.
解:(1)$(x + 2m)(x^{2} - x + \frac{1}{2}n)$
$ = x^{3} - x^{2} + \frac{1}{2}nx + 2mx^{2} - 2mx + mn$
$ = x^{3} + (2m - 1)x^{2} + (\frac{1}{2}n - 2m)x + mn$.
$\because$不含$x$项与$x^{2}$项,
$\therefore\begin{cases}2m - 1 = 0\\\frac{1}{2}n - 2m = 0\end{cases}$,解得$\begin{cases}m = \frac{1}{2}\\n = 2\end{cases}$.
(2)$m^{2023}n^{2024} = (\frac{1}{2})^{2023} \cdot 2^{2024} = (\frac{1}{2} \times 2)^{2023} \times 2 = 2$.
答案：
(1)$(x + 2m)(x^{2}-x+\frac{1}{2}n)$
$=x^{3}-x^{2}+\frac{1}{2}nx + 2mx^{2}-2mx + mn$
$=x^{3}+(2m - 1)x^{2}+(\frac{1}{2}n - 2m)x + mn$.
$\because$不含 $x$ 项与 $x^{2}$ 项，
$\therefore\begin{cases}2m - 1 = 0\\\frac{1}{2}n - 2m = 0\end{cases}$，解得$\begin{cases}m=\frac{1}{2}\\n = 2\end{cases}$.
(2)$m^{2023}n^{2024}=(\frac{1}{2})^{2023}\cdot2^{2024}=(\frac{1}{2}\times2)^{2023}\times2 = 2$.
(1)$(x + 2m)(x^{2}-x+\frac{1}{2}n)$
$=x^{3}-x^{2}+\frac{1}{2}nx + 2mx^{2}-2mx + mn$
$=x^{3}+(2m - 1)x^{2}+(\frac{1}{2}n - 2m)x + mn$.
$\because$不含 $x$ 项与 $x^{2}$ 项，
$\therefore\begin{cases}2m - 1 = 0\\\frac{1}{2}n - 2m = 0\end{cases}$，解得$\begin{cases}m=\frac{1}{2}\\n = 2\end{cases}$.
(2)$m^{2023}n^{2024}=(\frac{1}{2})^{2023}\cdot2^{2024}=(\frac{1}{2}\times2)^{2023}\times2 = 2$.
21. (4分)(1)在下列横线上用含有$a$,$b$的代数式表示相应图形的面积:
图(1):$a^{2}$;图(2)$2ab$;
图(3):$b^{2}$;图(4)$(a + b)^{2}$.
(2)通过拼图你发现前三个图形的面积之和与第四个正方形的面积之间有什么关系? 用数学式子表示为$(a + b)^{2} = a^{2} + 2ab + b^{2}$.
(3)利用(2)的结论计算$98^{2} + 392 + 4$的值.
解:(3)$98^{2} + 392 + 4 = 98^{2} + 2 \times 2 \times 98 + 2^{2} = (98 + 2)^{2} = 10000$.
图(1):$a^{2}$;图(2)$2ab$;
图(3):$b^{2}$;图(4)$(a + b)^{2}$.
(2)通过拼图你发现前三个图形的面积之和与第四个正方形的面积之间有什么关系? 用数学式子表示为$(a + b)^{2} = a^{2} + 2ab + b^{2}$.
(3)利用(2)的结论计算$98^{2} + 392 + 4$的值.
解:(3)$98^{2} + 392 + 4 = 98^{2} + 2 \times 2 \times 98 + 2^{2} = (98 + 2)^{2} = 10000$.
答案：
(1)$a^{2}$ $2ab$ $b^{2}$ $(a + b)^{2}$
(2)$(a + b)^{2}=a^{2}+2ab + b^{2}$
(3)$98^{2}+392 + 4 = 98^{2}+2\times2\times98 + 2^{2}=(98 + 2)^{2}=10000$.
(1)$a^{2}$ $2ab$ $b^{2}$ $(a + b)^{2}$
(2)$(a + b)^{2}=a^{2}+2ab + b^{2}$
(3)$98^{2}+392 + 4 = 98^{2}+2\times2\times98 + 2^{2}=(98 + 2)^{2}=10000$.
查看更多完整答案，请扫码查看
