分母有理化

分母有理化，簡稱有理化，指的是將該原為無理數的分母化為有理數的過程，也就是將分母中的根號化去。

有理化後通常方便運算，有理化的過程可能會影響分子，但分子及分母的比例不變。

單項式

應用一般根號運算：

${\displaystyle {\frac {1}{\sqrt {a}}}={\frac {1{\sqrt {a}}}{{\sqrt {a}}{\sqrt {a}}}}={\frac {\sqrt {a}}{a}}}$

${\displaystyle {\frac {1}{\sqrt[{n}]{a}}}={\frac {\sqrt[{n}]{a^{n-1}}}{a}}}$

二項式

應用平方差公式：

${\displaystyle {\frac {1}{{\sqrt {a}}+{\sqrt {b}}}}={\frac {{\sqrt {a}}-{\sqrt {b}}}{a-b}}}$

${\displaystyle {\frac {1}{{\sqrt {a}}-{\sqrt {b}}}}={\frac {{\sqrt {a}}+{\sqrt {b}}}{a-b}}}$

${\displaystyle {\frac {1}{{\sqrt {a}}+b}}={\frac {{\sqrt {a}}-b}{a-b^{2}}}}$

${\displaystyle {\frac {1}{{\sqrt {a}}-b}}={\frac {{\sqrt {a}}+b}{a-b^{2}}}}$

應用立方和、立方差公式：

${\displaystyle {\frac {1}{{\sqrt[{3}]{a}}+{\sqrt[{3}]{b}}}}={\frac {{\sqrt[{3}]{a^{2}}}-{\sqrt[{3}]{ab}}+{\sqrt[{3}]{b^{2}}}}{a+b}}}$

${\displaystyle {\frac {1}{{\sqrt[{3}]{a}}-{\sqrt[{3}]{b}}}}={\frac {{\sqrt[{3}]{a^{2}}}+{\sqrt[{3}]{ab}}+{\sqrt[{3}]{b^{2}}}}{a-b}}}$

${\displaystyle {\frac {1}{{\sqrt[{3}]{a}}+b}}={\frac {{\sqrt[{3}]{a^{2}}}-{\sqrt[{3}]{a}}b+b^{2}}{a+b^{3}}}}$

${\displaystyle {\frac {1}{{\sqrt[{3}]{a}}-b}}={\frac {{\sqrt[{3}]{a^{2}}}+{\sqrt[{3}]{a}}b+b^{2}}{a-b^{3}}}}$

多項式

逐項有理化

${\displaystyle {\frac {1}{{\sqrt {a}}+{\sqrt {b}}+c}}={\frac {{\sqrt {a}}-{\sqrt {b}}-c}{a-b-c^{2}-2c{\sqrt {b}}}}}$^[1]

輾轉相除法

設${\displaystyle x={\sqrt[{3}]{2}}}$有理化${\displaystyle {\frac {1}{1+2{\sqrt[{3}]{2}}+3{\sqrt[{3}]{4}}}}}$

${\displaystyle (x^{3}-2)u(x)+(1+2x+3x^{2})v(x)=1}$

${\displaystyle u(x)={\frac {-1}{89}}(50+3x),v(x)={\frac {1}{89}}(-11+16x+x^{2})}$

${\displaystyle {\frac {1}{1+2{\sqrt[{3}]{2}}+3{\sqrt[{3}]{4}}}}=v({\sqrt[{3}]{2}})={\frac {1}{89}}(-11+16{\sqrt[{3}]{2}}+{\sqrt[{3}]{4}})}$^[2]

待定係數法

${\displaystyle x^{3}=2x^{2}+3x+4}$，求${\displaystyle {\frac {1}{3+2x+x^{2}}}}$

設${\displaystyle (3+2x+x^{2})(a+bx+cx^{2})=1}$

${\displaystyle {\begin{pmatrix}1&x&x^{2}&x^{3}&x^{4}\end{pmatrix}}{\begin{pmatrix}3&0&0\\2&3&0\\1&2&3\\0&1&2\\0&0&1\end{pmatrix}}{\begin{pmatrix}a\\b\\c\end{pmatrix}}={\begin{pmatrix}1&x&x^{2}&x^{3}\end{pmatrix}}{\begin{pmatrix}3&0&0\\2&3&4\\1&2&6\\0&1&4\end{pmatrix}}{\begin{pmatrix}a\\b\\c\end{pmatrix}}={\begin{pmatrix}1&x&x^{2}\end{pmatrix}}{\begin{pmatrix}3&4&16\\2&6&16\\1&4&14\end{pmatrix}}{\begin{pmatrix}a\\b\\c\end{pmatrix}}={\begin{pmatrix}1&x&x^{2}\end{pmatrix}}{\begin{pmatrix}1\\0\\0\end{pmatrix}}}$

${\displaystyle {\begin{pmatrix}a\\b\\c\end{pmatrix}}={\begin{pmatrix}3&4&16\\2&6&16\\1&4&14\end{pmatrix}}^{-1}{\begin{pmatrix}1\\0\\0\end{pmatrix}}={\frac {1}{22}}{\begin{pmatrix}10\\-6\\1\end{pmatrix}}}$

${\displaystyle {\frac {1}{x^{2}+2x+3}}={\frac {x^{2}-6x+10}{22}}}$^[2]

參見

• 輾轉相除法