有双重根号的表示式在根号下还有根号,如: 1 + 2 + 3 5 {\displaystyle {\sqrt[{5}]{1+{\sqrt {2}}+{\sqrt {3}}}}} 在5次根号下有3个2次根号项。
如果m次根号内的表示式是由一个含根号的多项式自乘m次得来的,都可以化简。
a2-b为平方数时就可以化简 a ± b {\displaystyle {\sqrt {a\pm {\sqrt {b}}}}} 。
a ± b = a + a 2 − b ± a − a 2 − b 2 {\displaystyle {\sqrt {a\pm {\sqrt {b}}}}={\frac {{\sqrt {a+{\sqrt {a^{2}-b}}}}\pm {\sqrt {a-{\sqrt {a^{2}-b}}}}}{\sqrt {2}}}}
例如: 2 + 3 = 3 + 1 2 = 6 + 2 2 {\displaystyle {\sqrt {2+{\sqrt {3}}}}={\frac {{\sqrt {3}}+1}{\sqrt {2}}}={\frac {{\sqrt {6}}+{\sqrt {2}}}{2}}}
1 + 2 2 5 + 4 5 = ( 1 + 2 5 ) 2 = 1 + 2 5 {\displaystyle {\sqrt {1+2{\sqrt[{5}]{2}}+{\sqrt[{5}]{4}}}}={\sqrt {(1+{\sqrt[{5}]{2}})^{2}}}=1+{\sqrt[{5}]{2}}}
5 − 12 3 3 + 6 9 3 3 = ( 2 ) 3 − 3 ( 2 ) 2 3 3 + 3 ( 2 ) 9 3 − 3 3 = 2 − 3 3 {\displaystyle {\sqrt[{3}]{5-12{\sqrt[{3}]{3}}+6{\sqrt[{3}]{9}}}}={\sqrt[{3}]{(2)^{3}-3(2)^{2}{\sqrt[{3}]{3}}+3(2){\sqrt[{3}]{9}}-3}}=2-{\sqrt[{3}]{3}}}
对于 a ± b m {\displaystyle {\sqrt[{m}]{{\sqrt {a}}\pm {\sqrt {b}}}}} ,设 x 1 = a + b m , x 2 = a − b m {\displaystyle x_{1}={\sqrt[{m}]{{\sqrt {a}}+{\sqrt {b}}}},x_{2}={\sqrt[{m}]{{\sqrt {a}}-{\sqrt {b}}}}}
找x1+x2时需要用到 x 1 m + x 2 m = ∑ r = 0 ⌊ m 2 ⌋ m C m − r r m − r ( x 1 + x 2 ) m − 2 r ( − x 1 x 2 ) r {\displaystyle x_{1}^{m}+x_{2}^{m}=\sum _{r=0}^{\lfloor {\frac {m}{2}}\rfloor }{\frac {mC_{m-r}^{r}}{m-r}}(x_{1}+x_{2})^{m-2r}(-x_{1}x_{2})^{r}} [1]
x = x 1 + x 2 ± ( x 1 + x 2 ) 2 − 4 x 1 x 2 2 {\displaystyle x={\frac {x_{1}+x_{2}\pm {\sqrt {(x_{1}+x_{2})^{2}-4x_{1}x_{2}}}}{2}}}
27 − 28 3 {\displaystyle {\sqrt[{3}]{{\sqrt {27}}-{\sqrt {28}}}}}
对于 k 1 ± k 2 a ± k 3 b + k 4 a b m {\displaystyle {\sqrt[{m}]{k_{1}\pm k_{2}{\sqrt {a}}\pm k_{3}{\sqrt {b}}+k_{4}{\sqrt {ab}}}}} ,设 x 1 = k 1 + k 2 a + k 3 b + k 4 a b m , x 2 = k 1 − k 2 a − k 3 b + k 4 a b m {\displaystyle x_{1}={\sqrt[{m}]{k_{1}+k_{2}{\sqrt {a}}+k_{3}{\sqrt {b}}+k_{4}{\sqrt {ab}}}},x_{2}={\sqrt[{m}]{k_{1}-k_{2}{\sqrt {a}}-k_{3}{\sqrt {b}}+k_{4}{\sqrt {ab}}}}}
15 + 10 2 + 8 3 + 6 6 {\displaystyle {\sqrt {15+10{\sqrt {2}}+8{\sqrt {3}}+6{\sqrt {6}}}}}
55 + 81 2 2 + 33 3 + 45 2 6 3 {\displaystyle {\sqrt[{3}]{55+{\frac {81}{2}}{\sqrt {2}}+33{\sqrt {3}}+{\frac {45}{2}}{\sqrt {6}}}}}