查看“︁初中數學（香港課程）/多項式的因式/3”︁的源代码

小學的分數加減：
* <math>\frac{1}{5}+\frac{2}{5}=\frac{3}{5}</math>
* <math>\frac{3}{5}-\frac{1}{5}=\frac{2}{5}</math>

有相同分母代數分式的加減運算只需分子加減

{{MathHKQA|1=1
|2=化簡
# <math>\frac{9}{l-6}-\frac{4}{l-6}</math>
# <math>\frac{8w}{w+7}+\frac{56}{w+7}</math>
|sol=
# <math>\frac{9}{\color{red}l-6}-\frac{4}{\color{red}l-6}=\frac{5}{l-6}</math>
# <math>\begin{align}
\frac{8w}{w+7}+\frac{56}{w+7}&=\frac{8w+56}{w+7}\\
&=\frac{8(w+7)}{w+7}=8
\end{align}</math>
}}
小學的分數加法（通分母）：
* <math>\frac{1}{4}+\frac{1}{6}=\frac{3}{12}+\frac{2}{12}=\frac{5}{12}</math>

分母不同的分式需要把全部分式化為一樣分母，通常會化為全部分母的最少公倍數
{{MathHKQA|1=2
|2=化簡
# <math>\frac{3}{x}-\frac{1}{2x}</math>
# <math>\frac{4}{5b-6}+\frac{7}{20b-24}</math>
# <math>\frac{10}{n-8}+\frac{9}{8-n}</math>
|sol=
# <math>\begin{align}
\frac{3}{x}-\frac{1}{2x}&=\frac{3\cdot 2}{x\cdot 2}-\frac{1}{2x}\\
&=\frac{6}{2x}-\frac{1}{2x}=\frac{5}{2x}
\end{align}</math>
# <math>\begin{align}
\frac{4}{5b-6}+\frac{7}{20b-24}&=\frac{16}{4(5b-6)}+\frac{7}{4(5b-6)}\\
&=\frac{23}{4(5b-6)}
\end{align}</math>
# <math>\begin{align}
\frac{10}{n-8}+\frac{9}{8-n}&=\frac{10}{n-8}-\frac{9}{n-8}\\
&=\frac{1}{n-8}
\end{align}</math>
}}
{{MathHKQA|1=3
|2=化簡
# <math>\frac{4}{9a}+\frac{5}{6a}-\frac{7}{8a}</math>
# <math>\frac{1}{30y-20}-\frac{8}{18-27y}</math>
|sol=
# <math>\begin{align}
\frac{4}{9a}+\frac{5}{6a}-\frac{7}{8a}&=\frac{32}{72a}+\frac{60}{72a}-\frac{63}{72a}\\
&=\frac{29}{72a}
\end{align}</math>
# <math>\begin{align}
\frac{1}{30y-20}-\frac{8}{18-27y}&=\frac{1}{10(3y-2)}-\frac{8}{9(2-3y)}\\
&=\frac{1}{10(3y-2)}+\frac{8}{9(3y-2)}\\
&=\frac{9}{90(3y-2)}+\frac{80}{90(3y-2)}=\frac{89}{90(3y-2)}
\end{align}</math>
}}
{{MathHKQA|1=4
|2=化簡：
# <math>4-\frac{7}{m+1}</math>
# <math>\frac{3}{5z+1}+\frac{2}{z-6}</math>
|sol=
# <math>\begin{align}
4-\frac{7}{m+1}&=\frac{4(m+1)}{m+1}-\frac{7}{m+1}\\
&=\frac{4m+4}{m+1}-\frac{7}{m+1}=\frac{4m-3}{m+1}
\end{align}</math>
# <math>\begin{align}
\frac{3}{5z+1}+\frac{2}{z-6}&=\frac{3(z-6)}{(5z+1)(z-6)}+\frac{2(5z+1)}{(5z+1)(z-6)}\\
&=\frac{3z-18}{(5z+1)(z-6)}+\frac{10z+2}{(5z+1)(z-6)}\\
&=\frac{13z-16}{(5z+1)(z-6)}
\end{align}</math>
}}