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分母有理化 Connected to: {{::readMoreArticle.title}} 分母有理化,简称有理化,指的是将该原为无理数的分母化为有理数的过程,也就是将分母中的根号化去。 有理化后通常方便运算,有理化的过程可能會影響分子,但分子及分母的比例不變。 单项式[编辑]应用一般根号运算: 1 a = 1 a a a = a a {\displaystyle {\frac {1}{\sqrt {a))}={\frac {1{\sqrt {a))}((\sqrt {a)){\sqrt {a))))={\frac {\sqrt {a)){a))} 1 a n = a n − 1 n a {\displaystyle {\frac {1}{\sqrt[{n}]{a))}={\frac {\sqrt[{n}]{a^{n-1))}{a))} 二项式[编辑]应用平方差公式: 1 a + b = a − b a − b {\displaystyle {\frac {1}((\sqrt {a))+{\sqrt {b))))={\frac ((\sqrt {a))-{\sqrt {b))}{a-b))} 1 a − b = a + b a − b {\displaystyle {\frac {1}((\sqrt {a))-{\sqrt {b))))={\frac ((\sqrt {a))+{\sqrt {b))}{a-b))} 1 a + b = a − b a − b 2 {\displaystyle {\frac {1}((\sqrt {a))+b))={\frac ((\sqrt {a))-b}{a-b^{2)))) 1 a − b = 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: 1 a 3 + b 3 = a 2 3 − a b 3 + b 2 3 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))} 1 a 3 − b 3 = a 2 3 + a b 3 + b 2 3 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))} 1 a 3 + b = a 2 3 − a 3 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)))) 1 a 3 − b = a 2 3 + a 3 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)))) 多项式[编辑]逐项有理化[编辑] 1 a + b + c = a − b − c a − b − c 2 − 2 c b {\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[编辑]设 x = 2 3 {\displaystyle x={\sqrt[{3}]{2))} 有理化 1 1 + 2 2 3 + 3 4 3 {\displaystyle {\frac {1}{1+2{\sqrt[{3}]{2))+3{\sqrt[{3}]{4))))} ( x 3 − 2 ) u ( x ) + ( 1 + 2 x + 3 x 2 ) v ( x ) = 1 {\displaystyle (x^{3}-2)u(x)+(1+2x+3x^{2})v(x)=1} u ( x ) = − 1 89 ( 50 + 3 x ) , v ( x ) = 1 89 ( − 11 + 16 x + x 2 ) {\displaystyle u(x)={\frac {-1}{89))(50+3x),v(x)={\frac {1}{89))(-11+16x+x^{2})} 1 1 + 2 2 3 + 3 4 3 = v ( 2 3 ) = 1 89 ( − 11 + 16 2 3 + 4 3 ) {\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[编辑] x 3 = 2 x 2 + 3 x + 4 {\displaystyle x^{3}=2x^{2}+3x+4} ,求 1 3 + 2 x + x 2 {\displaystyle {\frac {1}{3+2x+x^{2)))) 设 ( 3 + 2 x + x 2 ) ( a + b x + c x 2 ) = 1 {\displaystyle (3+2x+x^{2})(a+bx+cx^{2})=1} ( 1 x x 2 x 3 x 4 ) ( 3 0 0 2 3 0 1 2 3 0 1 2 0 0 1 ) ( a b c ) = ( 1 x x 2 x 3 ) ( 3 0 0 2 3 4 1 2 6 0 1 4 ) ( a b c ) = ( 1 x x 2 ) ( 3 4 16 2 6 16 1 4 14 ) ( a b c ) = ( 1 x x 2 ) ( 1 0 0 ) {\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))} ( a b c ) = ( 3 4 16 2 6 16 1 4 14 ) − 1 ( 1 0 0 ) = 1 22 ( 10 − 6 1 ) {\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))} 1 x 2 + 2 x + 3 = x 2 − 6 x + 10 22 {\displaystyle {\frac {1}{x^{2}+2x+3))={\frac {x^{2}-6x+10}{22))} [2] 参见[编辑]辗转相除法参考资料[编辑] 数学主题 ^ 分母有理化与分子有理化. [2013-10-10]. (原始内容存档于2019-06-03). ^ 2.0 2.1 韩士安 林磊. 近世代数(第二版). 分类 分类:初等代数 {{bottomLinkPreText}} {{bottomLinkText}} This page is based on a Wikipedia article written by contributors (read/edit). Text is available under the CC BY-SA 4.0 license; additional terms may apply. Images, videos and audio are available under their respective licenses. Cover photo is available under {{::mainImage.info.license.name || 'Unknown'}} license. Cover photo is available under {{::mainImage.info.license.name || 'Unknown'}} license. Credit: (see original file).