分母有理化

应用一般根号运算：

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

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

应用平方差公式：

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

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

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

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

应用立方和、立方差公式：

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

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

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

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

逐项有理化

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

辗转相除法

设${\displaystyle x={\sqrt[{3}]{2}}}$ 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]

待定系数法

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

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

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

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

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

• 辗转相除法