10. (百色中考改编)已知$a = b + 2023$,则代数式$\dfrac{2}{a - b}·\dfrac{a^{2}-b^{2}}{a^{2}+2ab + b^{2}}÷\dfrac{1}{a^{2}-b^{2}}$的值为
4046
.
答案：10. 4046 解析：原式$=\frac {2}{a-b}· \frac {(a-b)(a+b)}{(a+b)^{2}}· (a-b)(a+b)=2(a-b)$，$\because a=b+2023$，$\therefore a-b=2023$，$\therefore$原式$=2×2023=4046$.
解析：
解：原式$=\dfrac{2}{a - b}·\dfrac{(a - b)(a + b)}{(a + b)^2}÷\dfrac{1}{a^2 - b^2}$
$=\dfrac{2}{a - b}·\dfrac{(a - b)(a + b)}{(a + b)^2}·(a^2 - b^2)$
$=\dfrac{2}{a - b}·\dfrac{(a - b)(a + b)}{(a + b)^2}·(a - b)(a + b)$
$=2(a - b)$
$\because a = b + 2023$
$\therefore a - b = 2023$
$\therefore$原式$=2×2023 = 4046$
4046
11. (1)已知$a^{4}+\dfrac{1}{a^{4}}=7$,则$a^{2}+\dfrac{1}{a^{2}}=$
3
.
(2)(广东中考)若$x+\dfrac{1}{x}=\dfrac{13}{6}$且$0 ＜ x ＜ 1$,则$x^{2}-\dfrac{1}{x^{2}}=$
$-\dfrac{65}{36}$
.
答案：11. (1)3 解析：$\because a^{4}+\frac {1}{a^{4}}=7$，$\therefore (a^{2}+\frac {1}{a^{2}})^{2}=9$，$\therefore a^{2}+\frac {1}{a^{2}}=3$或$a^{2}+\frac {1}{a^{2}}=-3$(舍去).
(2)$-\frac {65}{36}$ 解析：$\because 0＜x＜1$，$\therefore x＜\frac {1}{x}$，$\therefore x-\frac {1}{x}＜0$，$\because x+\frac {1}{x}=\frac {13}{6}$，$\therefore (x+\frac {1}{x})^{2}=\frac {169}{36}$，即$x^{2}+2+\frac {1}{x^{2}}=\frac {169}{36}$，$\therefore x^{2}-2+\frac {1}{x^{2}}=\frac {169}{36}-4$，$\therefore (x-\frac {1}{x})^{2}=\frac {25}{36}$，$\therefore x-\frac {1}{x}=-\frac {5}{6}$，$\therefore x^{2}-\frac {1}{x^{2}}=(x+\frac {1}{x})(x-\frac {1}{x})=\frac {13}{6}×(-\frac {5}{6})=-\frac {65}{36}$.
12. 新题型
新运算 我们定义一种新运算:记$a*b = (a + b)^{2}-(a - b)^{2}$,如果设$A$为代数式,$A*\dfrac{1}{x + 2y}=\dfrac{4}{x^{2}-4y^{2}}$,则$A =$
$\dfrac{1}{x-2y}$
(用含$x$,$y$的代数式表示).
答案：12. $\frac {1}{x-2y}$ 解析：$\because a*b=(a+b)^{2}-(a-b)^{2}=[(a+b)+(a-b)][(a+b)-(a-b)]=2a· 2b=4ab$，$\therefore A*\frac {1}{x+2y}=\frac {4}{x^{2}-4y^{2}}$可变形为$4A· \frac {1}{x+2y}=\frac {4}{(x+2y)(x-2y)}$，$\therefore A=\frac {4}{(x+2y)(x-2y)}· \frac {x+2y}{4}=\frac {1}{x-2y}$.
解析：
$\because a*b=(a+b)^{2}-(a-b)^{2}=[(a+b)+(a-b)][(a+b)-(a-b)]=2a·2b=4ab$，
$\therefore A*\frac{1}{x+2y}=4A·\frac{1}{x+2y}=\frac{4}{x^{2}-4y^{2}}$，
$\because x^{2}-4y^{2}=(x+2y)(x-2y)$，
$\therefore 4A·\frac{1}{x+2y}=\frac{4}{(x+2y)(x-2y)}$，
两边同时乘以$(x+2y)$得：$4A=\frac{4}{x-2y}$，
$\therefore A=\frac{1}{x-2y}$。
$\frac{1}{x-2y}$
13. 先化简,再求值:$(\dfrac{x^{2}-y^{2}}{xy})^{2}÷ (x + y)·(\dfrac{x}{x - y})^{3}$,其中$x = -\dfrac{1}{2}$,$y = -1$.
答案：13. 原式$=\frac {(x+y)^{2}(x-y)^{2}}{x^{2}y^{2}}· \frac {1}{x+y}· \frac {x^{3}}{(x-y)^{3}}=\frac {x(x+y)}{y^{2}(x-y)}$.当$x=-\frac {1}{2}$，$y=-1$时，原式$=\frac {3}{2}$.
解析：
解：原式$=\frac{(x+y)^{2}(x-y)^{2}}{x^{2}y^{2}}·\frac{1}{x+y}·\frac{x^{3}}{(x-y)^{3}}=\frac{x(x+y)}{y^{2}(x-y)}$.
当$x=-\frac{1}{2}$，$y=-1$时，
原式$=\frac{(-\frac{1}{2})×(-\frac{1}{2}+(-1))}{(-1)^{2}×(-\frac{1}{2}-(-1))}=\frac{(-\frac{1}{2})×(-\frac{3}{2})}{1×\frac{1}{2}}=\frac{\frac{3}{4}}{\frac{1}{2}}=\frac{3}{2}$.
14. 已知$A=\dfrac{x^{2}+x}{x - 4}÷\dfrac{x^{2}-1}{x^{2}-8x + 16}$,$B=\dfrac{x^{2}-x}{1 - x}$.
(1)化简分式$A$.
(2)若$C = A×\dfrac{1}{B}$,则当$x$取什么整数时,分式$C$的值为整数?
答案：14. (1)$A=\frac {x(x+1)}{x-4}÷\frac {(x+1)(x-1)}{(x-4)^{2}}=\frac {x(x+1)}{x-4}· \frac {(x-4)^{2}}{(x+1)(x-1)}=\frac {x^{2}-4x}{x-1}$.
(2)$C=A×\frac {1}{B}=\frac {x^{2}-4x}{x-1}· \frac {1-x}{x^{2}-x}=\frac {x-4}{-x-1}=-(1-\frac {3}{x-1})=\frac {3}{x-1}-1$，要使得分式C的值为整数，则x可取$-2$，0，2，4，$\because x-4≠0$，$(x+1)(x-1)≠0$，$x^{2}-8x+16≠0$，$1-x≠0$，$x^{2}-x≠0$，$\therefore x≠4$且$x≠\pm 1$且$x≠0$，$\therefore x$可取$-2$或2.
15. (武汉自主招生)已知$a_{1}=\dfrac{1}{2}$,$a_{2}=\dfrac{1}{3}+\dfrac{2}{3}$,$···$,$a_{n}=\dfrac{1}{n + 1}+\dfrac{2}{n + 1}+···+\dfrac{n}{n + 1}$,设$S_{n}=\dfrac{1}{a_{1}a_{2}}+\dfrac{1}{a_{2}a_{3}}+···+\dfrac{1}{a_{n}a_{n + 1}}$,则与$S_{2024}$最接近的整数为
4
.
答案：15. 4 解析：$a_{n}=\frac {1}{n+1}+\frac {2}{n+1}+... +\frac {n}{n+1}=\frac {1+2+... +n}{n+1}=\frac {\frac {n}{2}(n+1)}{n+1}=\frac {n}{2}$，$\therefore a_{n+1}=\frac {n+1}{2}$，$\therefore a_{n}a_{n+1}=\frac {n}{2}· \frac {n+1}{2}=\frac {n(n+1)}{4}$，$\therefore \frac {1}{a_{n}a_{n+1}}=\frac {4}{n(n+1)}$，$\therefore S_{n}=\frac {1}{a_{1}a_{2}}+\frac {1}{a_{2}a_{3}}+... +\frac {1}{a_{n}a_{n+1}}=4[\frac {1}{1×2}+\frac {1}{2×3}+... +\frac {1}{n(n+1)}]=4(1-\frac {1}{2}+\frac {1}{2}-\frac {1}{3}+... +\frac {1}{n}-\frac {1}{n+1})=4(1-\frac {1}{n+1})=\frac {4n}{n+1}$.当$n=2024$时，$S_{2024}=\frac {4×2024}{2025}\approx 4$.
解析：
$a_{n}=\frac{1}{n+1}+\frac{2}{n+1}+···+\frac{n}{n+1}=\frac{1+2+···+n}{n+1}=\frac{\frac{n(n+1)}{2}}{n+1}=\frac{n}{2}$，则$a_{n+1}=\frac{n+1}{2}$，
$a_{n}a_{n+1}=\frac{n}{2}·\frac{n+1}{2}=\frac{n(n+1)}{4}$，
$\frac{1}{a_{n}a_{n+1}}=\frac{4}{n(n+1)}=4(\frac{1}{n}-\frac{1}{n+1})$，
$S_{n}=4[(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+···+(\frac{1}{n}-\frac{1}{n+1})]=4(1-\frac{1}{n+1})=\frac{4n}{n+1}$，
当$n=2024$时，$S_{2024}=\frac{4×2024}{2025}\approx4$，
故与$S_{2024}$最接近的整数为$4$。
16. 某示范基地的“水稻1号”试验田是边长为$a$米($a ＞ 1$)的正方形去掉一个边长为$1$米的正方形蓄水池后余下的部分,“水稻2号”试验田是边长为$(a - 1)$米的正方形,两块试验田的水稻都收获了$1000$千克.
(1)试说明哪种水稻的单位面积产量高;
(2)高的单位面积产量是低的单位面积产量的多少倍?
答案：16. (1)$\because$“水稻1号”试验田是边长为a米的正方形减去一个边长为1米的正方形蓄水池后余下的部分，“水稻2号”试验田是边长为$(a-1)$米的正方形，$\therefore$“水稻1号”试验田的面积为$(a^{2}-1)$平方米，“水稻2号”试验田的面积为$(a-1)^{2}$平方米.
$\because a^{2}-1-(a-1)^{2}=a^{2}-1-a^{2}+2a-1=2(a-1)$，由题意可知，$a＞1$，$\therefore 2(a-1)＞0$，即$a^{2}-1＞(a-1)^{2}$.
$\because$两块试验田的水稻都收获了1000千克，$\therefore$“水稻2号”试验田的单位面积产量高.
(2)$\frac {1000}{(a-1)^{2}}÷\frac {1000}{a^{2}-1}=\frac {1000}{(a-1)^{2}}· \frac {(a+1)(a-1)}{1000}=\frac {a+1}{a-1}$.
答：高的单位面积产量是低的单位面积产量的$\frac {a+1}{a-1}$倍.
