初中數學（香港課程）/多項式的因式/3

小學的分數加減：

• $\frac{1}{5}+\frac{2}{5}=\frac{3}{5}$
• $\frac{3}{5}-\frac{1}{5}=\frac{2}{5}$

有相同分母代數分式的加減運算只需分子加減

示範例子1 化簡 $\frac{9}{l-6}-\frac{4}{l-6}$ $\frac{8w}{w+7}+\frac{56}{w+7}$ 解 $\frac{9}{l-6}-\frac{4}{l-6}=\frac{5}{l-6}$ $\begin{array}{cc}\frac{8w}{w+7}+\frac{56}{w+7}& =\frac{8w+56}{w+7}\\ & =\frac{8(w+7)}{w+7}=8\end{array}$

同類習題1 {{{3}}}

小學的分數加法（通分母）：

• $\frac{1}{4}+\frac{1}{6}=\frac{3}{12}+\frac{2}{12}=\frac{5}{12}$

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

示範例子2 化簡 $\frac{3}{x}-\frac{1}{2x}$ $\frac{4}{5b-6}+\frac{7}{20b-24}$ $\frac{10}{n-8}+\frac{9}{8-n}$ 解 $\begin{array}{cc}\frac{3}{x}-\frac{1}{2x}& =\frac{3\cdot 2}{x\cdot 2}-\frac{1}{2x}\\ & =\frac{6}{2x}-\frac{1}{2x}=\frac{5}{2x}\end{array}$ $\begin{array}{cc}\frac{4}{5b-6}+\frac{7}{20b-24}& =\frac{16}{4(5b-6)}+\frac{7}{4(5b-6)}\\ & =\frac{23}{4(5b-6)}\end{array}$ $\begin{array}{cc}\frac{10}{n-8}+\frac{9}{8-n}& =\frac{10}{n-8}-\frac{9}{n-8}\\ & =\frac{1}{n-8}\end{array}$

同類習題2 {{{3}}}

示範例子3 化簡 $\frac{4}{9a}+\frac{5}{6a}-\frac{7}{8a}$ $\frac{1}{30y-20}-\frac{8}{18-27y}$ 解 $\begin{array}{cc}\frac{4}{9a}+\frac{5}{6a}-\frac{7}{8a}& =\frac{32}{72a}+\frac{60}{72a}-\frac{63}{72a}\\ & =\frac{29}{72a}\end{array}$ $\begin{array}{cc}\frac{1}{30y-20}-\frac{8}{18-27y}& =\frac{1}{10(3y-2)}-\frac{8}{9(2-3y)}\\ & =\frac{1}{10(3y-2)}+\frac{8}{9(3y-2)}\\ & =\frac{9}{90(3y-2)}+\frac{80}{90(3y-2)}=\frac{89}{90(3y-2)}\end{array}$

同類習題3 {{{3}}}

示範例子4 化簡： $4-\frac{7}{m+1}$ $\frac{3}{5z+1}+\frac{2}{z-6}$ 解 $\begin{array}{cc}4-\frac{7}{m+1}& =\frac{4(m+1)}{m+1}-\frac{7}{m+1}\\ & =\frac{4m+4}{m+1}-\frac{7}{m+1}=\frac{4m-3}{m+1}\end{array}$ $\begin{array}{cc}\frac{3}{5z+1}+\frac{2}{z-6}& =\frac{3(z-6)}{(5z+1)(z-6)}+\frac{2(5z+1)}{(5z+1)(z-6)}\\ & =\frac{3z-18}{(5z+1)(z-6)}+\frac{10z+2}{(5z+1)(z-6)}\\ & =\frac{13z-16}{(5z+1)(z-6)}\end{array}$

同類習題4 {{{3}}}