初中数学（香港课程）/多项式的因式/2

由多项式P除以多项式Q组成不是多项式的代数式（${\frac {P}{Q}}$）称作代数分式（英：Algebraic Fraction）

• 若果Q是数字，则成为多项式（$Q\neq 0$），代数分式的分母不可以是数字
• 例子：${\frac {4}{x}}$、${\frac {a+1}{a-3}}$、 ${\frac {1}{p^{2}-5pq+q^{2}}}$

Extended content
${\frac {x}{9x}}$和${\frac {7a-56}{a-8}}$能否直接化简？如是，试化简 试因式分解${\frac {7a-56}{a-8}}$的分子，并化简

代数分式中分子和分母的公因式需因式分解才可取出，之后再消去公因式化简

示范例子1 化简： ${\frac {9b-54}{b^{2}-6b}}$ ${\frac {20y}{4y+8}}$ ${\frac {3c^{3}+30c^{2}d}{7c^{2}d+70cd^{2}}}$ 解 ${\begin{aligned}{\frac {9b-54}{b^{2}-6b}}&={\frac {9{\cancel {(b-6)}}}{b{\cancel {(b-6)}}}}\\&={\frac {9}{b}}\end{aligned}}$ ${\begin{aligned}{\frac {20y}{4y+8}}&={\frac {20y}{4(y+2)}}\\&={\frac {5y}{y+2}}\end{aligned}}$ ${\begin{aligned}{\frac {3c^{2}d+30cd^{2}}{7c^{3}+70c^{2}d}}&={\frac {{\color {red}3cd}{\color {blue}(c+10d)}}{{\color {red}7c^{2}}{\color {blue}(c+10d)}}}\\&={\frac {3d}{7c}}\end{aligned}}$

同类习题1 {{{3}}}

示范例子2 化简： ${\frac {8-z}{6z-48}}$ ${\frac {9r+45-rs-5s}{r+5}}$ 解 ${\begin{aligned}{\frac {8-z}{6z-48}}&={\frac {-(z-8)}{6(z-8)}}\\&=-{\frac {1}{6}}\end{aligned}}$ ${\begin{aligned}{\frac {s-9}{9r+45-rs-5s}}&={\frac {-(9-s)}{9(r+5)-s(r+5)}}\\&={\frac {-(9-s)}{(9-s)(r+5)}}\\&=-{\frac {1}{r+5}}\end{aligned}}$

同类习题2 {{{3}}}

乘法与除法

小学的分数乘法： ${\frac {5}{6}}\cdot {\frac {8}{15}}={\frac {{\cancel {5}}^{1}}{{\cancel {6}}_{3}}}\cdot {\frac {{\cancel {8}}^{4}}{{\cancel {15}}_{3}}}={\frac {4}{9}}$

代数分式的乘法与分数步骤相同，只需把两个分式的分子互乘、分母互乘即可，分子、分母之间可约简

• 化简仍需因式分解

小学的分数除法： ${\frac {\frac {1}{14}}{\frac {3}{7}}}={\frac {1}{14}}\cdot {\frac {7}{3}}={\frac {1}{6}}$

同样地，代数分式的除法中，分母和分子需倒转，化为乘法

示范例子3 化简： ${\frac {2k+18}{10}}\cdot {\frac {8}{k^{2}+9k}}$ ${\frac {\frac {21u^{5}}{6v^{4}}}{\frac {7u^{3}}{24v}}}$ ${\frac {\frac {4n+20}{n-1}}{\frac {3n+15}{n^{2}-n}}}$ 解 ${\begin{aligned}{\frac {2k+18}{10}}\cdot {\frac {8}{k^{2}+9k}}&={\frac {{\cancel {2}}{\cancel {(k+9)}}}{{\cancel {10}}_{5}}}\cdot {\frac {8}{k{\cancel {(k+9)}}}}\\&={\frac {8}{5k}}\end{aligned}}$ ${\begin{aligned}{\frac {\frac {21u^{5}}{6v}}{\frac {\color {red}7u^{3}}{\color {blue}24v^{4}}}}&={\frac {21u^{5}}{6v}}\cdot {\frac {\color {blue}24v^{4}}{\color {red}7u^{3}}}\\&={\frac {12u^{2}}{v^{3}}}\end{aligned}}$ ${\begin{aligned}{\frac {\frac {4n+20}{n-1}}{\frac {3n+15}{n^{2}-n}}}&={\frac {4n+20}{n-1}}\cdot {\frac {n^{2}-n}{3n+15}}\\&={\frac {4(n+5)}{n-1}}\cdot {\frac {n(n-1)}{3(n+5)}}\\&={\frac {4n}{3}}\end{aligned}}$

同类习题3 {{{3}}}

示范例子4 化简： ${\frac {56m-63m^{2}}{m^{3}}}\cdot {\frac {m^{5}}{42-54m}}$ ${\frac {{\frac {4h+8}{h^{6}}}\cdot h}{\frac {h^{2}+2h}{h^{3}}}}$ 解 ${\begin{aligned}{\frac {56m-72m^{2}}{m^{3}}}\cdot {\frac {m^{5}}{42-54m}}&={\frac {8m(7-9m)}{m^{3}}}\cdot {\frac {m^{5}}{6(7-9m)}}\\&={\frac {4}{3m^{3}}}\end{aligned}}$ ${\begin{aligned}{\frac {{\frac {4h+8}{h^{6}}}\cdot h}{\frac {h^{2}+2h}{h^{3}}}}&={\frac {4h+8}{h^{2}}}\cdot h^{6}\cdot {\frac {h^{3}}{h^{2}+2h}}&={\frac {4(h+2)}{h^{5}}}\cdot {\frac {h^{3}}{h(h+2)}}\\&={\frac {4}{h^{3}}}\end{aligned}}$

同类习题4 {{{3}}}