《2025年综合应用创新题典中点八年级数学上册冀教版》

13. (8分)计算：
(1)$\frac{x^2 - 4x + 4}{x^2 - 1}\cdot\frac{x + 1}{x^2 - 2x}+\frac{1}{x - 1}$；
(2)[2024泸州]$(\frac{y^2}{x}+x - 2y)÷\frac{x^2 - y^2}{x}$.

答案： 【解】
(1)原式$=\frac {(x-2)^{2}}{(x+1)(x-1)}\cdot \frac {x+1}{x(x-2)}+\frac {1}{x-1}$$=\frac {x-2}{x(x-1)}+\frac {1}{x-1}=\frac {x-2+x}{x(x-1)}=\frac {2}{x}.$
(2)原式$=\frac {y^{2}+x^{2}-2xy}{x}÷\frac {(x+y)(x-y)}{x}$$=\frac {(x-y)^{2}}{x}\cdot \frac {x}{(x+y)(x-y)}$$=\frac {x-y}{x+y}.$

14. (8分)先化简,再求值：$\frac{a}{a - b}÷\frac{a^2 - b^2}{a^2 - 2ab + b^2}-\frac{a - b}{a + b}$,其中$a,b满足b - 2a = 0$.

答案： 【解】原式$=\frac {a}{a-b}\cdot \frac {(a-b)^{2}}{(a+b)(a-b)}-\frac {a-b}{a+b}=\frac {a}{a+b}-$$\frac {a-b}{a+b}=\frac {b}{a+b}.$$\because b-2a=0,\therefore b=2a$.
∴原式$=\frac {2a}{a+2a}=\frac {2}{3}.$

15. (8分)已知$\frac{3x^2 - 7x + 2}{(x - 1)(x + 1)}= 3+\frac{a}{x - 1}+\frac{b}{x + 1}$是恒等式,请分别求$a,b$的值.

答案： 【解】因为$\frac {3x^{2}-7x+2}{(x-1)(x+1)}=3+\frac {a}{x-1}+\frac {b}{x+1}$是恒等式，且$3+\frac {a}{x-1}+\frac {b}{x+1}=\frac {3(x-1)(x+1)+a(x+1)+b(x-1)}{(x-1)(x+1)}=$$\frac {3x^{2}+(a+b)x+a-b-3}{(x-1)(x+1)}$,所以$3x^{2}-7x+2=3x^{2}+(a+b)x+a-b-3.$所以$\left\{\begin{array}{l} a+b=-7,\\ a-b-3=2,\end{array}\right. $解得$\left\{\begin{array}{l} a=-1,\\ b=-6.\end{array}\right. $

16. (10分)已知：$A = (\frac{x}{x - 1}-\frac{x}{x^2 - 1})÷\frac{x^2 - x}{x^2 - 2x + 1}$.
(1)化简$A$；
(2)若$x是不等式组\begin{cases}2x + 5\geq3\frac{1 - 3x}{2}\gt - 4\end{cases} $的整数解,求$A$的值.

答案： 【解】
(1)$A=\frac {x(x+1)-x}{(x+1)(x-1)}\cdot \frac {(x-1)^{2}}{x(x-1)}=\frac {x^{2}}{(x+1)(x-1)}\cdot$$\frac {x-1}{x}=\frac {x}{x+1}.$
(2)$\left\{\begin{array}{l} 2x+5≥3,①\\ \frac {1-3x}{2}＞-4,②\end{array}\right. $由①得$x≥-1$,由②得$x＜3,$
∴不等式组的解集为$-1≤x＜3.$
∴不等式组的整数解为-1,0,1,2.又$\because x+1≠0,x-1≠0,x≠0,\therefore x≠\pm 1$且$x≠0.$
∴当$x=2$时,$A=\frac {2}{3}.$

17. (14分)新考法 从特殊到一般的思想 观察下列式子：
$\frac{1}{2}= \frac{1}{1×2}= \frac{1}{1}-\frac{1}{2};\frac{1}{6}= \frac{1}{2×3}= \frac{1}{2}-\frac{1}{3};\frac{1}{12}= \frac{1}{3×4}= \frac{1}{3}-\frac{1}{4};\frac{1}{20}= \frac{1}{4×5}= \frac{1}{4}-\frac{1}{5};…$
(1)由此可以推测：$\frac{1}{42}=$$\frac{1}{6}-\frac{1}{7}$,$\frac{1}{72}=$$\frac{1}{8}-\frac{1}{9}$；
(2)请猜想出能表示出上述式子的特点的一般规律,用含字母$n$的等式表示出来($n$为正整数),并证明；
(3)请用(2)中的规律计算：$\frac{1}{(x - 2)(x - 3)}-\frac{2}{(x - 1)(x - 3)}+\frac{1}{(x - 1)(x - 2)}$.
(2)一般规律为$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$.
证明：$\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n(n+1)}-\frac{n}{n(n+1)}=\frac{1}{n(n+1)}$.
(3)原式$=\frac{1}{x-3}-\frac{1}{x-2}-(\frac{1}{x-3}-\frac{1}{x-1})+\frac{1}{x-2}-\frac{1}{x-1}$
$=\frac{1}{x-3}-\frac{1}{x-2}-\frac{1}{x-3}+\frac{1}{x-1}+\frac{1}{x-2}-\frac{1}{x-1}$
$=0$.

答案： 【解】
(1)$\frac {1}{6}-\frac {1}{7};\frac {1}{8}-\frac {1}{9}$
(2)一般规律为$\frac {1}{n(n+1)}=\frac {1}{n}-\frac {1}{n+1}.$证明:$\frac {1}{n}-\frac {1}{n+1}=\frac {n+1}{n(n+1)}-\frac {n}{n(n+1)}=\frac {1}{n(n+1)}.$
(3)原式$=\frac {1}{x-3}-\frac {1}{x-2}-(\frac {1}{x-3}-\frac {1}{x-1})+\frac {1}{x-2}-\frac {1}{x-1}$$=\frac {1}{x-3}-\frac {1}{x-2}-\frac {1}{x-3}+\frac {1}{x-1}+\frac {1}{x-2}-\frac {1}{x-1}$$=0.$