香港中學會考 - 2006 - 卷二 -

答案：A
$\begin{array}{rcl}
\dfrac{1}{x+1} – \dfrac

答案：D
$\begin{array}{cl}
& \dfrac{1}{n+3} – \

答案：A
$\begin{array}{rcl}
\dfrac{-k}{1-k} – \dfra

答案：D
$\begin{array}{cl}
& \dfrac{1}{a-2} – \

答案：D
$\begin{array}{cl}
& \dfrac{1}{2x-3} + \dfrac

答案：B
設 $2^p{a}^q{b}^r{c}^s$ 為第 3 個數式。

$\begin{array}{rlllllll}

答案：B
$\begin{array}{rcl}
a^2+4a+4 & = & (a+2)^

答案：A
$\begin{array}{lllllllllll}
3x^4y^2z & = 

答案：C
$\begin{array}{cl}
& \dfrac{1}{x^2 -2x +1} 

答案：${\displaystyle \frac{-13}{(4x-5)(1-6x)}}$
$\begin{array}{cl}
&#

答案：C
$\begin{array}{lcccccccccc}
9a^2b & = & &

答案：A
$\begin{array}{cl}
& \dfrac{1}{3x + 7} –

答案：${\displaystyle \frac{x}{(7x-6)(5x-4)}}$
$\begin{array}{cl}
&

答案：A
$\begin{array}{rcl}
\dfrac{5}{4k+3} – \dfra

答案：B

$\begin{array}{cl}
& \dfrac{6}{n-6} – \

答案：A

$\begin{array}{rcl}
\text{第 $1$ 個數式} & = &

答案：${\displaystyle \frac{17k}{(k-9)(5k+6)}}$

$\begin{array}{cl}
�

答案：C

$\begin{array}{cl}
& \dfrac{2}{5-4x}-\dfrac

答案：D

在該三個數式中，$a$、$b$ 及 $c$ 的最高次數分別為 $4$、$5$ 及 $2$。所以，該 L.C