8. 已知$\frac{x}{x^{2}+1}=\frac{1}{3}$，求$\frac{x^{2}}{x^{4}+x^{2}+1}$的值.
答案：8．由$\frac{x}{x^2 + 1} = \frac{1}{3}$，知$x \neq 0$，$\therefore \frac{x^2 + 1}{x} = 3$，即$x + \frac{1}{x} = 3$．两边同
时平方，得$x^2 + 2 + \frac{1}{x^2} = 9$，即$x^2 + \frac{1}{x^2} = 7$．$\therefore \frac{x^4 + x^2 + 1}{x^2} = x^2 +$
$1 + \frac{1}{x^2} = 8$．$\therefore \frac{x^2}{x^4 + x^2 + 1} = \frac{1}{8}$
9. 已知$\frac{
x}{2}=\frac{
y}{7}=\frac{
z}{5}$，则$\frac{x + y - z}{x}$的值为(
C
)
A.$\frac{1}{2}$
B.$-\frac{1}{2}$
C.2
D.$-2$
答案：9．C
解析：
设$\frac{x}{2}=\frac{y}{7}=\frac{z}{5}=k$，则$x = 2k$，$y = 7k$，$z = 5k$。
$\frac{x + y - z}{x}=\frac{2k + 7k - 5k}{2k}=\frac{4k}{2k}=2$
答案：C
10. 若$a:b=\frac{9}{25}:0.54$，$a:c=\frac{4}{5}:1\frac{1}{3}$，则$\frac{a + 3b - 2c}{b + c - a}$的值为
1
.
答案：10．1
解析：
$a:b=\frac{9}{25}:0.54=2:3$，设$a=2k$，则$b=3k$；$a:c=\frac{4}{5}:1\frac{1}{3}=3:5$，则$c=\frac{10}{3}k$。
$\frac{a + 3b - 2c}{b + c - a}=\frac{2k + 3×3k - 2×\frac{10}{3}k}{3k + \frac{10}{3}k - 2k}=\frac{2k + 9k - \frac{20}{3}k}{\frac{9}{3}k + \frac{10}{3}k - \frac{6}{3}k}=\frac{\frac{6}{3}k + \frac{27}{3}k - \frac{20}{3}k}{\frac{13}{3}k}=\frac{\frac{13}{3}k}{\frac{13}{3}k}=1$
1
11. 已知$a$，$b$，$c$满足$\frac{a}{2}=\frac{b - c}{3}=\frac{a + c}{5}$，求$\frac{a + c}{2a + b}$的值.
答案：11．设$\frac{a}{2} = \frac{b - c}{3} = \frac{a + c}{5} = k (k \neq 0)$，则$a = 2k$①，$b - c = 3k$②，
$a + c = 5k$③．①$+$②$+$③，得$2a + b = 10k$．$\therefore \frac{a + c}{2a + b} = \frac{5k}{10k} = \frac{1}{2}$
12. 已知$\frac{y + z}{x}=\frac{x + z}{y}=\frac{x + y}{z}$，其中$x + y + z\neq0$，求$\frac{x + y - z}{x + y + z}$的值.
答案：12．由题意，得$x$，$y$，$z$均不为$0$．设$\frac{y + z}{x} = \frac{x + z}{y} = \frac{x + y}{z} =$
$k (k \neq 0)$，则$\begin{cases} y + z = kx ①, \\ x + z = ky ②, \\ x + y = kz ③. \end{cases}$①$+$②$+$③，得$2x + 2y + 2z =$
$k (x + y + z)$．$\because x + y + z \neq 0$，$\therefore k = 2$．$\therefore \frac{x + y}{z} = 2$，即$x + y =$
$2z$．$\therefore$原式$= \frac{2z - z}{2z + z} = \frac{z}{3z} = \frac{1}{3}$
