《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. (湖南长郡中学自主招生)如果$a+\frac{1}{b}=1,b+\frac{2}{c}=1$,那么$c+\frac{2}{a}$等于 (A)

A.2
B.1
C.4
D.3

答案： 1.A 提示:$\because a + \frac{1}{b} = 1,\therefore a = \frac{b - 1}{b}$,又$\because b + \frac{2}{c} = 1,\therefore c = \frac{2}{1 - b}$,$\therefore c + \frac{2}{a} = \frac{2}{1 - b} + \frac{2b}{b - 1} = \frac{2 - 2b}{1 - b} = 2$. 故选 A.

2. (安徽芜湖一中自主招生)已知非零实数$a,b,c$满足$\frac{a^{2}}{1 + 2a^{2}}=\frac{b}{4},\frac{b^{2}}{4 + 3b^{2}}=\frac{c}{12},\frac{c^{2}}{4 + 6c^{2}}=\frac{a}{2}$,则$a + b + c=$ (A)

A.$\frac{11}{3}$
B.$\frac{13}{3}$
C.$\frac{17}{5}$
D.$\frac{23}{7}$

答案： 2.A 提示:$\because \frac{a^{2}}{1 + 2a^{2}} = \frac{b}{4},\frac{b^{2}}{4 + 3b^{2}} = \frac{c}{12},\frac{c^{2}}{4 + 6c^{2}} = \frac{a}{2},\therefore \frac{12}{a} + 2 = \frac{4}{b} + 3 = \frac{2}{c} + 6,\therefore \frac{4}{b} + \frac{12}{c} + \frac{2}{a} = \frac{1}{a^{2}} + \frac{4}{b^{2}} + \frac{c^{2}}{11}$,
$\therefore (\frac{1}{a^{2}} - \frac{2}{a} + 1) + (\frac{4}{b^{2}} - \frac{4}{b} + 1) + (\frac{12}{c^{2}} + 9) = 0$,即$(\frac{1}{a} - 1)^{2} + (\frac{2}{b} - 1)^{2} + (\frac{2}{c} - 3)^{2} = 0,\therefore \frac{1}{a} = 1$,即$a = 1,\frac{2}{b} = 1$,即$b = 2$,
$\frac{2}{c} = 3$,即$c = \frac{2}{3},\therefore a + b + c = \frac{11}{3}$. 故选 A.

3. (北京市竞赛)当$m = -\frac{1}{6}$时,代数式$\frac{21 - 5m}{m^{2} - 9}-\frac{m}{m^{2} - 9}÷\frac{m}{m + 3}-\frac{m - 3}{m + 3}$的值是 (A)

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

答案： 3.A 提示:原式$= \frac{21 - 5m}{m^{2} - 9} - \frac{m + 3}{m^{2} - 9} + \frac{m - 3}{m + 3} = \frac{21 - 5m - m - 3 - m^{2} + 6m - 9}{m^{2} - 9} = \frac{9 - m^{2}}{m^{2} - 9} = - 1$. 故选 A.

4. (2022·广东省深圳市第九届“鹏程杯”八年级邀请赛)一列数$a_{1},a_{2},a_{3},·s,a_{n}$,其中$a_{1} = -1,a_{2} = \frac{1}{1 - a_{1}},a_{3} = \frac{1}{1 - a_{2}},·s,a_{n} = \frac{1}{1 - a_{n - 1}}$,则$a_{1}× a_{2}× a_{3}×·s× a_{2022}=$ (B)

A.$-1$
B.1
C.2022
D.$-2$
E.$\frac{1}{2}$

答案： 4.B 提示:$\because a_{1} = - 1,a_{2} = \frac{1}{1 - a_{1}},·s,a_{n} = \frac{1}{1 - a_{n - 1}}$,$\therefore a_{2} = \frac{1}{1 - ( - 1)} = \frac{1}{2},a_{3} = \frac{1}{1 - \frac{1}{2}} = 2,a_{4} = \frac{1}{1 - 2} = - 1,·s$这列数以$- 1,\frac{1}{2},2$每三个数为一个周期依次循环,$\because 2022 ÷ 3 = 674$,$a_{1} × a_{2} × a_{3} = - 1 × \frac{1}{2} × 2 = - 1$,$\therefore a_{1} × a_{2} × a_{3} × ·s × a_{2022} = ( - 1)^{674} =$
1. 故选 B.

5. (太原市初中数学竞赛)已知$\frac{ab}{a + b}=\frac{1}{15},\frac{bc}{b + c}=\frac{1}{17},\frac{ca}{c + a}=\frac{1}{16}$,则$\frac{abc}{ab + bc + ba}$的值是 (D)

A.$\frac{1}{21}$
B.$\frac{1}{22}$
C.$\frac{1}{23}$
D.$\frac{1}{24}$

答案： 5.D 提示:由$\frac{ab}{a + b} = \frac{1}{15}$得,$\frac{a + b}{ab} = 15$,即$\frac{1}{a} + \frac{1}{b} = 15$,同理$\frac{1}{b} + \frac{1}{c} = 17$,$\frac{1}{a} + \frac{1}{c} = 16$,则有$2(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) = 15 + 17 + 16$,$\therefore \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 24$,故$\frac{abc}{ab + bc + ca} = \frac{1}{24}$. 故选 D.

6. (2022·广东省深圳市第九届“鹏程杯”八年级邀请赛)已知实数$x,y,z$满足$\frac{x}{y + z}+\frac{y}{x + z}+\frac{z}{x + y}=1$,则分式$\frac{x^{2}}{y + z}+\frac{y^{2}}{x + z}+\frac{z^{2}}{x + y}$的值为 (B)

A.$-1$
B.0
C.1
D.2
E.$-2$

答案： 6.B 提示:$\because$实数$x,y,z$满足$\frac{x}{y + z} + \frac{y}{x + z} + \frac{z}{x + y} = 1$,变形等式得$\frac{x}{y + z} = 1 - \frac{y}{x + z} - \frac{z}{x + y}$,$\frac{x^{2}}{y + z} = x - \frac{xy}{x + z} - \frac{xz}{x + y}$,同理可得$\frac{y^{2}}{x + z} = y - \frac{xy}{y + z} - \frac{xz}{y + x}$,
$\frac{z^{2}}{x + y} = z - \frac{yz}{y + z} - \frac{xz}{x + z}$,$\therefore \frac{x^{2}}{y + z} + \frac{y^{2}}{x + z} + \frac{z^{2}}{x + y} = x + y + z - (\frac{xy + yz}{y + z} + \frac{xy + xz}{x + z} + \frac{yz + xz}{x + y}) = x + y + z - (\frac{xy + yz + xy + xz + yz + xz}{x + y + z}) = 0$. 故选 B.

7. 如果$\frac{|a|}{a}+\frac{|b|}{b}+\frac{|c|}{c}=1$,则$\frac{|abc|}{abc}$的值为 (A)

A.$-1$
B.1
C.$\pm1$
D.不确定

答案： 7.A 提示:$\because \frac{|a|}{a} + \frac{|b|}{b} + \frac{|c|}{c} = 1$,则$a,b,c$为两正一负,不妨设$a ＞ 0,b ＞ 0,c ＜ 0$,则原式$= \frac{|abc|}{abc} = \frac{- abc}{abc} = - 1$. 故选 A.

8. (安徽省合肥自主招生)设$a,b,c$满足$abc\neq0$,且$a + b = c$,则$\frac{b^{2} + c^{2} - a^{2}}{2bc}+\frac{c^{2} + a^{2} - b^{2}}{2ca}+\frac{a^{2} + b^{2} - c^{2}}{2ab}$的值为 (B)

A.$-1$
B.1
C.2
D.3

答案： 8.B 提示:$\because a + b = c,\therefore b = c - a,c = a + b,a = c - b$,原式$= \frac{b^{2} + c(c - a)}{2bc} + \frac{c^{2} + a(a - b)}{2ca} + \frac{a^{2} + b(b - c)}{2ab}$
$= \frac{b^{2} + c^{2} - ac + c^{2} + a^{2} - ab + a^{2} + b^{2} - bc}{2abc} × 2ab$（此处原解析存在表达不清晰，正确推导应为：原式$= \frac{b^{2} + c(c - a)}{2bc} + \frac{c^{2} + a(a - b)}{2ca} + \frac{a^{2} + b(b - c)}{2ab} = \frac{b - b - c + c + a - 2b}{2b}$（此步错误，重新推导）
原式$= \frac{b^{2} + c^{2} - ac}{2bc} + \frac{c^{2} + a^{2} - ab}{2ca} + \frac{a^{2} + b^{2} - bc}{2ab} = 1 + 1 - 1 = 1$. 故选 B.

9. (南昌市竞赛)已知实数$p,q$满足条件:$\frac{1}{p}-\frac{1}{q}=\frac{1}{p + q}$,则代数式$\frac{q}{p}+\frac{p}{q}$的值为±√5.

答案： 9.$\pm \sqrt{5}$ 提示:$\because \frac{1}{p} + \frac{1}{p + q} + \frac{1}{p + q} = 1,\frac{q}{p} - \frac{p}{q} = 1,\therefore \frac{q}{p} + \frac{p}{q} + 4 = 5,\therefore \frac{q}{p} + \frac{p}{q} = \pm \sqrt{5}$.