分母有理化

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

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]

${\displaystyle x^{3}=2x^{2}+3x+4}$，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]