《2025年全国重点高中提前招生同步强化全真试卷八年级数学上册》

1. (五城市联赛试题)已知一列数$a_{1},a_{2},a_{3},a_{4},a_{5},a_{6},a_{7}$,且$a_{1}=8,a_{7}=5832,\frac {a_{1}}{a_{2}}=\frac {a_{2}}{a_{3}}=\frac {a_{3}}{a_{4}}=\frac {a_{4}}{a_{5}}=\frac {a_{5}}{a_{6}}=\frac {a_{6}}{a_{7}}$,则$a_{5}$为 (A)

A.648
B.832
C.1168
D.1944

答案： 1.A 提示$:k^{6}=\frac{a_{1}}{a_{2}} · \frac{a_{2}}{a_{3}} · \frac{a_{3}}{a_{4}} · \frac{a_{4}}{a_{5}} · \frac{a_{5}}{a_{6}} · \frac{a_{6}}{a_{7}}=\frac{a_{1}}{a_{7}}=\frac{8}{5832},$得$k= \pm \frac{1}{3} ,$又$\frac{a_{1}}{a_{5}}=\frac{a_{1}}{a_{2}} · \frac{a_{2}}{a_{3}} · \frac{a_{3}}{a_{4}} · \frac{a_{4}}{a_{5}}=k^{4}=\frac{1}{81} ,\because a_{5}=81a_{1}=81 × 8=648. $
故选A.

2. (2023·湖北黄冈中学自主招生)下列变形,正确的有 (B)
①若$ma^{2}=na^{2}$,则$m=n$;②若$\frac {r}{m}=\frac {r}{n}$,则$\frac {1}{m}=\frac {1}{n}$;③若$\frac {x}{a^{2}+1}=\frac {y}{a^{2}+1}$,则$x=y$;④若$\frac {a}{c}=\frac {b}{c}$,则$a=b$.

A.1个
B.2个
C.3个
D.4个

答案： 2.B 提示$:\textcircled{1} ma^{2}=na^{2},$若a=0,m和n不一定相等，故$\textcircled{1}$不符合题意$;\textcircled{2} \frac{r}{m}=\frac{r}{n},$若r=0,则$\frac{1}{m}$和$\frac{1}{n}$不一定相等，故$\textcircled{2}$不符合题意$;\textcircled{3}$若$\frac{x}{a^{2}+1}=\frac{y}{a^{2}+1},$则x=y,正确,故$\textcircled{3}$符合题意$;\textcircled{4}$若$\frac{a}{c}=\frac{b}{c},$则a=b,正确,故$\textcircled{4}$符合题意$. \therefore $正确的有2个.故选B.

3. (全国初中数学联赛试题)若$\frac {x}{3y}=\frac {y}{2x - 5y}=\frac {6x - 15y}{x}$,则$\frac {4x^{2}-5xy + 6y^{2}}{x^{2}-2xy + 3y^{2}}$的值为 (A)

A.$\frac {9}{2}$
B.$\frac {9}{4}$
C.5
D.6

答案： 3.A

4. (湖北武汉汉阳区自主招生)已知$abc = 1,a + b + c = 2,a^{2}+b^{2}+c^{2}=3$,则$\frac {1}{ab + c - 1}+\frac {1}{bc + a - 1}+\frac {1}{ca + b - 1}$的值为 (D)

A.-1
B.$-\frac {1}{2}$
C.2
D.$-\frac {2}{3}$

答案： 4.D 提示:由a+b+c=2,两边平方,得$a^{2}+b^{2}+c^{2}+2ab+2bc+2ac=4,$将已知代人,得$ab+bc+ac=\frac{1}{2};$由a+b+c=2得$:c-1=1-a-b,\therefore ab+c-1=ab+1-a-b=(a-1)(b-1),$同理,得$bc+a-1=(b-1)(c-1),ca+b-1=(c-1)(a-1),\therefore$原式$=\frac{1}{(a-1)(b-1)}+\frac{1}{(b-1)(c-1)}+\frac{1}{(c-1)(a-1)}=\frac{c-1+a-1+b-1}{(a-1)(b-1)(c-1)}=\frac{-1}{(ab-a-b+1)(c-1)}=abc-ac-bc+c-ab+a+b-1=\frac{-1}{1-\frac{1}{2}+2-1}=-\frac{2}{3}.$故选D.

5. (2023秋·相城区校级期末)已知实数$a,b,c$均不为零,且满足$a + b + c = 0$,则$\frac {1}{b^{2}+c^{2}-a^{2}}+\frac {1}{c^{2}+a^{2}-b^{2}}+\frac {1}{a^{2}+b^{2}-c^{2}}$的值 (C)

A.为正
B.为负
C.为0
D.与$a,b,c$的取值有关

答案： 5.C 提示$:\because a+b+c=0,\therefore b+c=-a,a+c=-b,a+b=-c,\therefore \frac{b^{2}+c^{2}-a^{2}}{2}+\frac{c^{2}+a^{2}-b^{2}}{2}+\frac{a^{2}+b^{2}-c^{2}}{2}=(b+c)^{2}-2bc-a^{2}+(c+a)^{2}-2ac-b^{2}+(a+b)^{2}-2ab-c^{2}=\frac{1}{2}[a^{2}-2bc-a^{2}+a^{2}-2ac-a^{2}+b^{2}-2ac-b^{2}+b^{2}-2ab-b^{2}+c^{2}-2ab-c^{2}]=\frac{1}{2} × [\frac{-1}{abc}(ab+bc+ca-a-b-c)]=\frac{1}{2} × \frac{a+b+c}{abc}=0.$故选C.

6. 已知三个数$x,y,z$满足$\frac {xy}{x + y}=-2,\frac {yz}{y + z}=\frac {4}{3},\frac {zx}{z + x}=-\frac {4}{3}$,则$\frac {xyz}{xy + yz + zx}$的值为 (D)

A.-2
B.$-\frac {1}{2}$
C.$-\frac {1}{4}$
D.-4

答案： 6.D 提示$:\because \frac{xy}{x+y}=-2,\frac{yz}{y+z}=\frac{4}{3},\frac{zx}{z+x}=-\frac{4}{3},\therefore \frac{x+y}{xy}=-\frac{1}{2},\frac{y+z}{yz}=\frac{3}{4},\frac{z+x}{zx}=-\frac{3}{4},\therefore \frac{x+y}{xy}+\frac{y+z}{yz}+\frac{z+x}{zx}=-\frac{1}{2}+\frac{3}{4}-\frac{3}{4}=-\frac{1}{2},\frac{1}{y}+\frac{1}{x}+\frac{1}{z}+\frac{1}{y}+\frac{1}{x}+\frac{1}{z}=-\frac{1}{2},\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2} × (-\frac{1}{2}-\frac{3}{4}-\frac{3}{4})=- \frac{1}{4} × \frac{xyz}{xyz+yz+zx}=-4.$故选D.

7. (湖北麻城市一中预录考试)已知$f(x)=\frac {x}{1 + x}$,则代数式$f(\frac {1}{2023})+f(\frac {1}{2022})+·s +f(\frac {1}{2})+f(1)+f(0)+f(1)+f(2)+·s +f(2022)+f(2023)$的值是 (D)

A.2020
B.2021
C.2022
D.2023

答案： 7.D 提示$:f(x)+f(\frac{1}{x})=\frac{x}{1+x}+\frac{1}{1+\frac{1}{x}}=\frac{x}{1+x}+\frac{x}{x+1}=1.\therefore$原式$=f(\frac{1}{2023})+f(2023)+f(\frac{1}{2022})+f(2022)+·s+f(0)=2023.$故选D.

8. (全国初中数学联赛)若$a,b$是两个正数,且$\frac {a - 1}{b}+\frac {b - 1}{a}+1 = 0$,则 (C)

A.$0\lt a + b\leqslant\frac {1}{3}$
B.$\frac {1}{3}\lt a + b\leqslant1$
C.$1\lt a + b\leqslant\frac {4}{3}$
D.$0\lt a + b\leqslant\frac {4}{3}$

答案： 8.C 提示:由$\frac{a^{2}-1}{b}+\frac{b-1}{a}+1=0,$可得$a^{2}-a+b^{2}-b=-ab,$则$ab=(a+b)^{2}-(a+b)=(a+b)(a+b-1).\textcircled{1}$由于a,b是两个正数$,\therefore ab＞0,a+b＞0,\therefore a+b-1＞0.\therefore a+b＞1,$另一个方面,由于$(a+b)^{2}=(a-b)^{2}+4ab \geqslant 4ab,\therefore ab \leqslant \frac{(a+b)^{2}}{4},$综合$\textcircled{1}$式,可得$\frac{a+b}{4} \geqslant a+b-1,\therefore a+b \leqslant \frac{4}{3},\therefore 1＜a+b \leqslant \frac{4}{3}.$故选C.

9. (2023·鼓楼区校级自主招生改编)已知$n,k$均为正整数,且对于每一个确定的$n$,满足不等式$\frac {5}{9}\lt\frac {n}{n + k}\lt\frac {4}{7}$的$k$仅有一个,则$n$的最大值与最小值的积为360.

答案： 9.360 提示$:\because \frac{5}{9} ＜ \frac{n}{n+k} ＜ \frac{4}{7},\therefore 5(n+k)＜9n,7n＜4(n+k),\therefore \frac{3}{4}n＜k＜\frac{4}{5}n,$即$\frac{15n}{20}＜k＜\frac{16n}{20},\because k$有唯一正整数解$,\therefore 9 \leqslant n \leqslant 40,\therefore n$的最大值为40,最小值为$9.\therefore 40 × 9=360.$