《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》

16. (北京市竞赛)设$x=\frac{a - b}{a + b},y=\frac{b - c}{b + c},z=\frac{c - a}{c + a}$,求证:$(1 + x)(1 + y)(1 + z)=(1 - x)(1 - y)(1 - z)$.

答案： 16.证明:$1 + x = \frac{a + b + a - b}{a + b} = \frac{2a}{a + b} - 1,1 - x = \frac{a + b - a + b}{a + b} = \frac{2b}{a + b}$,同理$1 + y = \frac{2b}{b + c},1 - y = \frac{2c}{b + c}$,$1 + z = \frac{2c}{c + a},1 - z = \frac{2a}{c + a}$. 因此$(1 + x)(1 + y)(1 + z) = \frac{8abc}{(a + b)(b + c)(c + a)} = (1 - x)(1 - y)(1 - z)$. 故有$(1 + x)(1 + y)(1 + z) = (1 - x)(1 - y)(1 - z)$.

17. (2023·福州鼓楼区自主招生)已知$a + b + c = 2023,\frac{a}{x^{2} - yz}=\frac{b}{y^{2} - xz}=\frac{c}{z^{2} - xy},x + y + z\neq0$,求$\frac{ax + by + cz}{x + y + z}$的值.

答案： 17.解:由题意,设$\frac{a}{x^{2} - yz} = \frac{b}{y^{2} - xz} = \frac{c}{z^{2} - xy} = k$,$\therefore a = k(x^{2} - yz),b = k(y^{2} - xz),c = k(z^{2} - xy)$. 原式$= \frac{kx(x^{2} - yz) + by(y^{2} - xz) + cz(z^{2} - xy)}{x + y + z} = \frac{k[(x^{3} + y^{3} + z^{3}) - 3xyz]}{x + y + z} = k\frac{(x + y + z)(x^{2} + y^{2} + z^{2} - xy - yz - xz)}{x + y + z} = k(x^{2} - yz) + k(y^{2} - xz) + k(z^{2} - xy) = a + b + c = 2023$.

18. (广东省深圳市第六届“鹏程杯”八年级邀请赛)若$abc = 1$,解方程$\frac{2ax}{ab + a + 1}+\frac{2bx}{bc + b + 1}+\frac{2cx}{ca + c + 1}=1$.

答案： 18.解:因为$abc = 1$,所以原方程可变形为$\frac{2ax}{ab + a + abc} + \frac{2bx}{bc + b + 1} + \frac{2cx}{ac + c + 1} = 1$,化简整理为$\frac{2(b + 1)x + 2bcx}{b(ac + c + 1)} = 1,\frac{2x(b + abc + bc)}{abc + bc + b} = 1,\therefore x = \frac{1}{2}$为原方程的解.