《2025年全效学习阶段发展评价九年级数学全一册浙教版》

7. 已知$\frac{x}{2} = \frac{y}{3} = \frac{z}{4} \neq 0$,则$\frac{x^{2} + xy}{yz} = $$\frac{5}{6}$.

答案： $\frac{5}{6}$　　[解析]设$\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=k$ (k≠0),　则x = 2k,y = 3k,z = 4k,
∴$\frac{x^{2}+xy}{yz}=\frac{(2k)^{2}+2k\cdot3k}{3k\cdot4k}=\frac{5}{6}$.

8. 已知$\frac{b}{a} = \frac{d}{c} \neq 1$,求证：$\frac{a + b}{a - b} = \frac{c + d}{c - d}$.

答案： 证明:
∵$\frac{b}{a}=\frac{d}{c}$,
∴1 + $\frac{b}{a}$ = 1 + $\frac{d}{c}$,
∴$\frac{a + b}{a}=\frac{c + d}{c}$.①
∵$\frac{b}{a}=\frac{d}{c}$≠1,
∴1 - $\frac{b}{a}$ = 1 - $\frac{d}{c}$≠0,
∴$\frac{a - b}{a}=\frac{c - d}{c}$≠0.②　①÷②,得$\frac{a + b}{a - b}=\frac{c + d}{c - d}$.

9. 已知$\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = \frac{2}{5}$.
(1)求$\frac{a - c}{b - d}$的值.
(2)求$\frac{2a + 3c - 4e}{2b + 3d - 4f}$的值.
(3)通过(1),(2)的计算,请你直接写出$\frac{la + mc + ne}{lb + md + nf}$的值($l,m,n$均不为 0).

答案： 解:
(1)由题意,得a = $\frac{2}{5}$b,c = $\frac{2}{5}$d,　则$\frac{a - c}{b - d}=\frac{\frac{2}{5}b-\frac{2}{5}d}{b - d}=\frac{2}{5}$.　
(2)由题意,得a = $\frac{2}{5}$b,c = $\frac{2}{5}$d,e = $\frac{2}{5}$f,　则$\frac{2a + 3c - 4e}{2b + 3d - 4f}=\frac{2×\frac{2}{5}b + 3×\frac{2}{5}d - 4×\frac{2}{5}f}{2b + 3d - 4f}=\frac{\frac{2}{5}(2b + 3d - 4f)}{2b + 3d - 4f}=\frac{2}{5}$.　
(3)$\frac{la + mc + ne}{lb + md + nf}=\frac{2}{5}$.

10. [原创]若$\frac{x + 1}{2} = \frac{y}{3} = \frac{z + 5}{4}$,且$3x - 2y + z = 24$,求$\frac{x + y}{z}$的值.

答案： 解:设$\frac{x + 1}{2}=\frac{y}{3}=\frac{z + 5}{4}=k$,　则x = 2k - 1,y = 3k,z = 4k - 5,
∴3(2k - 1) - 2·3k + (4k - 5)= 24,
∴k = 8,
∴x = 15,y = 24,z = 27,
∴$\frac{x + y}{z}=\frac{15 + 24}{27}=\frac{13}{9}$.

11. 若$\frac{a + b}{c} = \frac{b + c}{a} = \frac{c + a}{b} = k$,求$k$的值.

答案： 解:由题意,得a + b = ck,①　　b + c = ak,②　　c + a = bk.③　　① + ② + ③,得2(a + b + c)= k(a + b + c).　　分两种情况讨论:　　I.当a + b + c≠0时,k = 2.　　II.当a + b + c = 0时,a + b = -c,b + c = -a,a + c = -b,
∴k = -1.　　综上所述，k的值为2或 - 1.