《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》

10. (2024·重庆外国语学校自主招生改编)甲、乙相距300km,一辆汽车从甲地到乙地,如果车速提高20%,可提前1h到达,如果原速行驶$a$km后,再将速度提高25%,也可提前1h到达.则$a$是50km.

答案： 10.50 提示:设原来的车速是$x km/h$,$300 ÷ [x × (1 + 20\%)] = 300 ÷ x - 1$,解得$x = 50$;依题意得$\frac{a}{50} + \frac{300 - a}{50 × (1 + 25\%)} = \frac{300}{50} - 1$,整理得$62.5a + 15000 - 50a = 15625$,解得$a = 50$;经检验,$a = 50$是方程的根.

11. 已知$a\gt0,b\gt0,a + 2b = 1$,则$\frac{b^{2} + a + 1}{2ab}$的最小值为3+√10.

答案： 11.$3 + \sqrt{10}$ 提示:由题可得$\frac{b^{2} + a + 1}{2ab} = \frac{b^{2} + a(a + 2b) + (a + 2b)^{2}}{2ab} = \frac{5b^{2} + 6ab + 2a^{2}}{2ab} = \frac{5b}{2a} + \frac{a}{b} + 3 \geq 2\sqrt{\frac{5b}{2a} · \frac{a}{b}} + 3 = \sqrt{10} + 3$,当且仅当$\frac{5b}{2a} = \frac{a}{b}$即$a^{2} = 5b^{2}$时,等号成立.$\frac{b^{2} + a + 1}{2ab}$的最小值为$3 + \sqrt{10}$.

12. (2022·深圳罗湖区自主招生)已知$a^{2} + b^{2} + c^{2}=(a + b + c)^{2}$,则$\frac{b + c}{a}+\frac{c + a}{b}+\frac{a + b}{c}+100=$97.

答案： 12.97 提示:$\because a^{2} + b^{2} + c^{2} = (a + b + c)^{2} - 2(ab + bc + ac)$,$\therefore a^{2} + b^{2} + c^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ac$,$\therefore ab + bc + ac = 0$,$\therefore ab + bc + ac = 0$,$\therefore \frac{b + c}{a} + \frac{c + a}{b} + \frac{a + b}{c} + 100 = \frac{ - ac}{ac} + \frac{ - ac}{ac} + \frac{ - bc}{bc} + 100 = \frac{ - ac - ab}{ac} + \frac{ - ab - bc}{ab} + \frac{ - bc - ac}{bc} + 100 = - 1 - 1 - 1 + 100 = 97$.

13. (全国初中数学竞赛)如果$a,b,c$是正数,且满足$a + b + c = 9,\frac{1}{a + b}+\frac{1}{b + c}+\frac{1}{c + a}=\frac{10}{9}$,那么$\frac{a}{b + c}+\frac{b}{c + a}+\frac{c}{a + b}$的值为7.

答案： 13.7 提示:在$\frac{1}{a + b} + \frac{1}{b + c} + \frac{1}{c + a} = \frac{10}{9}$,左边乘以$a + b + c$,右边乘以$9$得$3 + \frac{c}{a + b} + \frac{a}{b + c} + \frac{b}{c + a} = 10$,得$\frac{c}{a + b} + \frac{a}{b + c} + \frac{b}{c + a} = 7$.

14. (2023·湖北黄冈中学自主招生)已知$\frac{5}{x^{2} + 2x}-x^{2}=2x + 4$,则代数式$5 - 2x^{2} - 4x$的值是3.

答案： 14.3 提示:令$t = x^{2} + 2x$,$t = x^{2} + 2x + 1 - 1 = (x + 1)^{2} - 1 \geq - 1$,$\therefore 5 - 2(x^{2} + 2x) = 5 - 2t = 5 - 2 = 3$.

15. (1)(温江区自主招生)给出以下式子:$(\frac{x^{2} - 4}{x^{2} - 4x + 4}-\frac{1}{x - 2})÷\frac{x + 1}{x + 2}$,先化简,然后从$-1,2,4 + 2\sqrt{3}$三个数中,选个合适的数代入求值.
(2)(华中师大一附中自主招生)已知$9a^{2} + 6ab + b^{2}=0$,求$\frac{a - 2b}{a^{2} + 3ab}÷\frac{1}{a^{2} - 9b^{2}}÷\frac{a^{2} - 4ab + 3b^{2}}{a}$的值.

答案： 15.解:
(1)$(\frac{x^{2} - 4}{x^{2} - 4x + 4} ÷ \frac{x + 1}{x - 2} - \frac{1}{x + 1}) ÷ \frac{x + 2}{x^{2} - 2} = [\frac{(x + 2)(x - 2)}{(x - 2)^{2}} - \frac{1}{x + 1}] ÷ \frac{x + 2}{x^{2} - 2} = (\frac{x + 2}{x - 2} - \frac{1}{x + 1}) ÷ \frac{x + 2}{x^{2} - 2} = \frac{x^{2} + 3x}{x^{2} - x - 2} ÷ \frac{x + 2}{x^{2} - 2} = \frac{x(x + 3)}{(x - 2)(x + 1)} · \frac{x^{2} - 2}{x + 2} = \frac{x(x + 3)(x^{2} - 2)}{(x - 2)(x + 1)(x + 2)}$
由题意得,$x + 1 \neq 0,x + 2 \neq 0,x - 2 \neq 0,x \neq 0$,$\therefore x \neq - 1,x \neq - 2,x \neq 2,x \neq 0$,$\therefore x = 4 + 2\sqrt{3}$,当$x = 4 + 2\sqrt{3}$时,原式$= \frac{4 + 2\sqrt{3}}{4 + 2\sqrt{3} - 2} · \frac{\sqrt{3}(1 + \sqrt{3})}{1 + \sqrt{3}} = \sqrt{3}$.
(2)$\because 9a^{2} + 6ab + b^{2} = 0,\therefore (3a + b)^{2} = 0$,$\therefore 3a + b = 0$. 原式$= \frac{a - 2b}{a^{2} + 3ab} · (a^{2} - 9b^{2}) = \frac{a - 2b}{a(a + 3b)} · (a + 3b)(a - 3b) · \frac{a}{(a - 3b)(a - b)} = \frac{a - 2b}{a} = \frac{7}{4}$.