1. (2025·南充)已知$\frac{a}{bc}=\frac{b}{ac}=\frac{c}{ab}=2$，则$\frac{a^{2}+b^{2}+c^{2}}{abc}$的值是(
D
)
A.2
B.3
C.4
D.6
答案：1．D
解析：
由$\frac{a}{bc}=2$，得$a = 2bc$；由$\frac{b}{ac}=2$，得$b = 2ac$；由$\frac{c}{ab}=2$，得$c = 2ab$。
将$a = 2bc$代入$b = 2ac$，得$b = 2(2bc)c = 4bc^2$，$b \neq 0$，两边同除以$b$，$1 = 4c^2$，$c^2=\frac{1}{4}$，同理$a^2=\frac{1}{4}$，$b^2=\frac{1}{4}$。
$a = 2bc$，则$a^2 = 4b^2c^2$，$\frac{1}{4}=4×\frac{1}{4}× c^2$，$c^2=\frac{1}{4}$，同理$abc=\frac{1}{8}$（$a,b,c$同号）。
$\frac{a^2 + b^2 + c^2}{abc}=\frac{\frac{1}{4}+\frac{1}{4}+\frac{1}{4}}{\frac{1}{8}}=\frac{\frac{3}{4}}{\frac{1}{8}}=6$
D
2. (2025·青海)先化简$(1-\frac{a}{a + 2})÷\frac{2}{a^{2}-4}$，再从$-2$，$0$，$1$中选一个合适的数作为$a$的值代入求值.
答案：2．原式$= ( \frac{a + 2}{a + 2} - \frac{a}{a + 2} ) · \frac{a^2 - 4}{2} = \frac{2}{a + 2} · \frac{(a + 2)(a - 2)}{2} =$
$a - 2$．由分母不能为$0$，得$a \neq \pm 2$．若选$0$作为$a$的值，则原式$=$
$0 - 2 = - 2$；若选$1$作为$a$的值，则原式$= 1 - 2 = - 1$
3. 若$x+\frac{1}{x}=\frac{13}{6}$，且$0\lt x\lt1$，则$x^{2}-\frac{1}{x^{2}}$的值为
$-\frac{65}{36}$
.
答案：3．$- \frac{65}{36}$ 解析：$\because 0 ＜ x ＜ 1$，$\therefore x ＜ \frac{1}{x}$，$\therefore x - \frac{1}{x} ＜ 0$．$\because x +$
$\frac{1}{x} = \frac{13}{6}$，$\therefore ( x + \frac{1}{x} )^2 = \frac{169}{36}$，即$x^2 + 2 + \frac{1}{x^2} = \frac{169}{36}$．$\therefore x^2 - 2 +$
$\frac{1}{x^2} = \frac{25}{36} - 4$．$\therefore ( x - \frac{1}{x} )^2 = \frac{25}{36}$．$\therefore x - \frac{1}{x} = - \frac{5}{6}$．$\therefore x^2 -$
$\frac{1}{x^2} = ( x + \frac{1}{x} )( x - \frac{1}{x} ) = \frac{13}{6} × ( - \frac{5}{6} ) = - \frac{65}{36}$．
4. 已知$y\gt x\gt0$，$x^{2}+y^{2}=4xy$，求$\frac{x + y}{x - y}$的值.
答案：4．$\because x^2 + y^2 = 4xy$，$\therefore x^2 + 2xy + y^2 = 6xy$，$x^2 - 2xy + y^2 =$
$2xy$．$\therefore (x + y)^2 = 6xy$，$(x - y)^2 = 2xy$．$\therefore \frac{(x + y)^2}{(x - y)^2} = \frac{6xy}{2xy}$，即
$( \frac{x + y}{x - y} )^2 = 3$．$\because y ＞ x ＞ 0$，$\therefore \frac{x + y}{x - y} ＜ 0$．$\therefore \frac{x + y}{x - y} = - \sqrt{3}$
5. 若实数$x$满足$x^{2}+3x - 1 = 0$，则$\frac{x}{x^{2}-1}$的值为
$-\frac{1}{3}$
.
答案：5．$- \frac{1}{3}$
解析：
解：由$x^{2}+3x - 1 = 0$，得$x^{2}-1=-3x$。
则$\frac{x}{x^{2}-1}=\frac{x}{-3x}=-\frac{1}{3}$。
$-\frac{1}{3}$
6. (2024·北京)先化简，再求值：$\frac{3(a - 2b)+3b}{a^{2}-2ab + b^{2}}$，其中$a$，$b$满足$a - b - 1 = 0$.
答案：6．原式$= \frac{3}{a - b}$．$\because a - b - 1 = 0$，$\therefore a - b = 1$．$\therefore$原式$= 3$
解析：
解：原式$=\frac{3a - 6b + 3b}{(a - b)^2}=\frac{3a - 3b}{(a - b)^2}=\frac{3(a - b)}{(a - b)^2}=\frac{3}{a - b}$。
$\because a - b - 1 = 0$，$\therefore a - b = 1$。
$\therefore$原式$=\frac{3}{1}=3$。
7. 已知$\frac{1}{a}+\frac{1}{b}=\frac{1}{2}$，$\frac{1}{b}+\frac{1}{c}=\frac{1}{3}$，$\frac{1}{a}+\frac{1}{c}=\frac{1}{4}$，求$\frac{abc}{ab + bc + ac}$的值.
答案：7．$\because \frac{1}{a} + \frac{1}{b} = \frac{1}{2}$，$\frac{1}{b} + \frac{1}{c} = \frac{1}{3}$，$\therefore \frac{1}{a} + \frac{1}{c} = \frac{1}{4}$，$\therefore \frac{1}{a} + \frac{1}{b} +$
$\frac{1}{b} + \frac{1}{c} + \frac{1}{a} + \frac{1}{c} = \frac{1}{2} + \frac{1}{3} + \frac{1}{4}$，即$\frac{2}{a} + \frac{2}{b} + \frac{2}{c} = \frac{13}{12}$．
$\therefore \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{13}{24}$．根据题意，得$abc \neq 0$，$\therefore \frac{abc}{ab + bc + ac} =$
$\frac{abc ÷ abc}{(ab + bc + ac) ÷ abc} = \frac{1}{\frac{1}{c} + \frac{1}{a} + \frac{1}{b}} = \frac{24}{13}$
