《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》

9. (2024·浙江省宁波市鄞州中学强基招生)已知$x,y,z$是大于$1$的正整数,且$(x+\frac{1}{y})(y+\frac{1}{z})(z+\frac{1}{x})$为整数,则$x + y + z=$12。

答案： 9.12 提示: $(x + \frac{1}{y})(y + \frac{1}{z})(z + \frac{1}{x}) = xyz + x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{xyz} ＜ 2$, 所以$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{xyz} = 1$, 若$x \geq 4$, 则$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{xyz} \leq \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{64} ＜ 1$不成立, 所以$x = 2$或3, 当$x = 2$时, $\frac{1}{y} + \frac{1}{z} + \frac{1}{2yz} = \frac{1}{2}$, 变形得$(y - 2)(z - 2) = 5 × 1$, 所以$y = 3$, $z = 7$, 则$x + y + z = 12$. 当$x = 3$时, $\frac{1}{y} + \frac{1}{z} + \frac{1}{3yz} = \frac{2}{3}$, 变形得$(2y - 3)(2z - 3) = 11 × 1$, 所以$y = 2$, $z = 7$, $y ＜ x$, 舍去. 综上, $x + y + z = 12$.

10. (2022·北师大台州附中学科类特长生考试)设$a,b,c,d$都是正数,且$S=\frac{a}{a + b + d}+\frac{b}{a + b + c}+\frac{c}{b + c + d}+\frac{d}{a + c + d}$,那么$S$的取值范围是1 ＜ S ＜ 2。

答案： 10.$1 ＜ S ＜ 2$ 提示: 因为$a$, $b$, $c$, $d$都为正数, 所以$S = \frac{a}{a + b + d} + \frac{b}{a + b + c} + \frac{c}{b + c + d} + \frac{d}{a + c + d} ＞ \frac{a}{a + b + c + d} + \frac{b}{a + b + c + d} + \frac{c}{a + b + c + d} + \frac{d}{a + b + c + d} = \frac{2(a + b + c + d)}{a + b + c + d} = 2$, 所以$1 ＜ S ＜ 2$.

11. (成都七中自主招生)方程组$\begin{cases}\frac{1}{x}+\frac{1}{y + z}=\frac{1}{3},\frac{1}{y}+\frac{1}{x + z}=\frac{1}{4},\frac{1}{z}+\frac{1}{x + y}=\frac{1}{5}\end{cases}$的解是 ______ 。

答案： 11.$\begin{cases} x = \frac{11}{3} \\ y = - \frac{11}{2} \\ z = 11 \end{cases}$ 提示: $\begin{cases} \frac{1}{y} + \frac{1}{x} = \frac{1}{3} - \textcircled{1} \\ \frac{1}{z} + \frac{1}{x} = \frac{1}{4} - \textcircled{2} \\ \frac{1}{z} + \frac{1}{y} = \frac{1}{5} - \textcircled{3} \end{cases}$原方程组可化为$\begin{cases} yz + xy = 4(x + y + z) - \textcircled{2} \\ xz - yz = x + y + z - \textcircled{4} \\ xz + yz = 5(x + y + z) - \textcircled{3} \end{cases}$ $\frac{5(yz - xz)}{(a + b)(a + c)} = \frac{2(c - b)}{(a + c)(a + c)} = \frac{2(a - c)}{(a + b)(b + c)} = \frac{8(a - c)(b - a)(c - b)}{(a + b)^2(b + c)^2(a + c)}$, 5(yz - xz) = 4yz,
∴ $6xz = 4yz$, 显然$z \neq 0$,
∴ $x = \frac{2}{3}y$ ⑤, ③ - ②得$xz - xy = x + y + z$ ⑥, 代入①得$xy + xz = 3(x + y + z)$,
∴ $4xy = 2xz$,
∴ $x \neq 0$, $z = 2y$ ⑦, 把⑤, ⑦分别代入①得$\frac{2}{3}y^2 + \frac{4}{3}y^2 = 3(\frac{2}{3}y + y + 2y)$,
∴ $2y^2 = 11y$,
∵ $y \neq 0$,
∴ $y = \frac{11}{2}$,
∴ $x = \frac{11}{3}$, $z = 2y = 11$,
∴ $\begin{cases} x = \frac{11}{3} \\ y = - \frac{11}{2} \\ z = 11 \end{cases}$

12. (2023·上海中学自主招生)已知$a+\frac{1}{b}=3,b+\frac{1}{c}=17,c+\frac{1}{a}=\frac{11}{25}$,则$abc=$1。

答案： 12.1 提示: 将三个条件式进行相乘, 得$(a + \frac{1}{b})(b + \frac{1}{c})(c + \frac{1}{a}) = 3 × 17 × \frac{11}{25}$, 整理得$(ab + \frac{a}{c} + 1 + \frac{1}{bc})(bc + \frac{b}{a} + 1 + \frac{1}{ac}) = abc + b + a + \frac{1}{c} + \frac{1}{b} + \frac{1}{abc} = abc + \frac{1}{abc} + 3 + 17 + \frac{11}{25} = abc + \frac{1}{abc} + \frac{511}{25} = 3 × 17 × \frac{11}{25}$,
∴ $abc + \frac{1}{abc} = 2$, 整理得$(abc)^2 - 2abc + 1 = 0$, 解得$abc = 1$.

13. (上海竞赛)设正整数$m,n$,满足$m\lt n$,且$\frac{1}{m^{2}+m}+\frac{1}{(m + 1)^{2}+(m + 1)}+·s+\frac{1}{n^{2}+n}=\frac{1}{23}$,则$m + n$的值是527。

答案： 13.527 提示:
∵ $\frac{1}{n^2 + n} = \frac{1}{n} - \frac{1}{n + 1}$,
∴ $\frac{1}{1^2 + 1} + \frac{1}{2^2 + 2} + \frac{1}{m^2 + m} + \frac{1}{(m + 1)^2 + (m + 1)} + ·s + \frac{1}{n^2 + n} = \frac{1}{1 × 2} + \frac{1}{2 × 3} + \frac{1}{m(m + 1)} + \frac{1}{(m + 1)(m + 2)} + ·s + \frac{1}{n(n + 1)} = \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{m} - \frac{1}{m + 1} + \frac{1}{m + 1} - \frac{1}{m + 2} + ·s + \frac{1}{n} - \frac{1}{n + 1} = \frac{23}{22} - \frac{1}{22 × 23} = \frac{23 - 1}{22 × 23} = \frac{22}{22 × 23} = \frac{1}{23}$,
∴ $m = 22$, $n + 1 = 23 × 22$,
∴ $n + 1 = 506$, $n = 505$, $m + n = 527$.

14. (浙江温州自主招生改编)若正实数$a,b,c$满足$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{2025}{a + b + c}$,则$\frac{(a + b)(b + c)(c + a)}{abc}$的值为2024。

答案： 14.2024 提示:
∵ $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{bc + ac + ab}{abc} = \frac{2025}{abc}$,
∴ $(a + b + c)(bc + ac + ab) = 2025abc$, 展开得$a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 3abc = 2025abc$,
∴ $(a + b)(b + c)(c + a) = a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 2abc$,
∴ $(a + b)(b + c)(c + a) = 2024abc$,
∴ $\frac{(a + b)(b + c)(c + a)}{abc} = 2024$.

15. (四川成都九中树德中学自主招生)已知$a^{2}+4a + 1 = 0$,且$\frac{a^{4}+ma^{2}+1}{3a^{3}+ma^{2}+3a}=5$,求$m$的值.

答案： 15.解:
∵ $a^2 + 4a + 1 = 0$,
∴ $a^2 = -4a - 1$,
∴ $\frac{a^4 + ma^2 + 1}{3a^3 + ma^2 + 3a} = \frac{(-4a - 1)^2 + ma^2 + 1}{3a(-4a - 1) + ma^2 + 3a} = \frac{(16 + m)a^2 + 8a + 2}{(m - 12)a^2} = \frac{(16 + m)(-4a - 1) + 8a + 2}{(m - 12)(-4a - 1)} = 5$,
∴ $(16 + m)(-4a - 1) + 8a + 2 = 5(m - 12)(-4a - 1)$, 原式可化为$(16 + m)(-4a - 1) - 5(m - 12)(-4a - 1) = -8a - 2$, 即$[(16 + m) - 5(m - 12)](-4a - 1) = -8a - 2$,
∵ $a \neq 0$,
∴ $4a + 1 \neq 0$,
∴ $(16 + m) - 5(m - 12) = 2$, 解得$m = \frac{37}{2}$.

16. (湖南长沙一中理科实验班招生)已知$x,y,z,a,b,c$均为实数,且$x = by + cz,y = cz + ax,z = ax + by$(其中$abcxyz\neq0$).求$\frac{1 - a}{1 + a}+\frac{1 - b}{1 + b}+\frac{1 - c}{1 + c}$的值.

答案： 16.解: 由题意得$ax + by + cz = \frac{x + y + z}{2}$, $ax + x = \frac{x + y + z}{2}$, $x(1 + a) = \frac{x + y + z}{2}$, $1 + a = \frac{x + y + z}{2x}$, 同理可得$1 + b = \frac{x + y + z}{2y}$, $1 + c = \frac{x + y + z}{2z}$, 原式$= \frac{2 - (1 + a)}{1 + a} + \frac{2 - (1 + b)}{1 + b} + \frac{2 - (1 + c)}{1 + c} = \frac{1 - a}{1 + a} + \frac{1 - b}{1 + b} + \frac{1 - c}{1 + c} = \frac{4x}{x + y + z} + \frac{4y}{x + y + z} + \frac{4z}{x + y + z} - 3 = \frac{4(x + y + z)}{x + y + z} - 3 = 4 - 3 = 1$.