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。 公式 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}}} 配方法 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}}} 增乘法 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}}}}} x 1 x 2 = 27 − 28 3 = − 1 {\displaystyle x_{1}x_{2}={\sqrt[{3}]{27-28}}=-1} ( x 1 + x 2 ) 3 + 3 ( x 1 + x 2 ) = 6 3 , x 1 + x 2 = 3 u , 3 u 3 + 3 u = 6 , u = 1 , x 1 + x 2 = 3 {\displaystyle (x_{1}+x_{2})^{3}+3(x_{1}+x_{2})=6{\sqrt {3}},x_{1}+x_{2}={\sqrt {3}}u,3u^{3}+3u=6,u=1,x_{1}+x_{2}={\sqrt {3}}} 27 − 28 3 = 3 − 7 2 {\displaystyle {\sqrt[{3}]{{\sqrt {27}}-{\sqrt {28}}}}={\frac {{\sqrt {3}}-{\sqrt {7}}}{2}}} 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}}}}} x 1 x 2 = 49 + 20 6 = 5 + 2 6 {\displaystyle x_{1}x_{2}={\sqrt {49+20{\sqrt {6}}}}=5+2{\sqrt {6}}} x 1 + x 2 = 40 + 16 6 = 4 + 2 6 {\displaystyle x_{1}+x_{2}={\sqrt {40+16{\sqrt {6}}}}=4+2{\sqrt {6}}} 15 + 10 2 + 8 3 + 6 6 = 4 + 2 6 + 20 + 8 6 2 = 2 + 2 + 3 + 6 {\displaystyle {\sqrt {15+10{\sqrt {2}}+8{\sqrt {3}}+6{\sqrt {6}}}}={\frac {4+2{\sqrt {6}}+{\sqrt {20+8{\sqrt {6}}}}}{2}}=2+{\sqrt {2}}+{\sqrt {3}}+{\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}}}}} x 1 x 2 = − 485 + 198 6 3 = − 5 − 2 6 {\displaystyle x_{1}x_{2}=-{\sqrt[{3}]{485+198{\sqrt {6}}}}=-5-2{\sqrt {6}}} ( x 1 + x 2 ) 3 + 3 ( 5 + 2 6 ) ( x 1 + x 2 ) = 5 ( 22 + 9 6 ) , ( 22 − 9 6 ) ( x 1 + x 2 ) 3 + 3 ( 2 − 6 ) ( x 1 + x 2 ) = − 10 , x 1 + x 2 = ( 2 + 6 ) u {\displaystyle (x_{1}+x_{2})^{3}+3(5+2{\sqrt {6}})(x_{1}+x_{2})=5(22+9{\sqrt {6}}),(22-9{\sqrt {6}})(x_{1}+x_{2})^{3}+3(2-{\sqrt {6}})(x_{1}+x_{2})=-10,x_{1}+x_{2}=(2+{\sqrt {6}})u} − 4 u 3 − 6 u = − 10 , u = 1 , x 1 + x 2 = 2 + 6 {\displaystyle -4u^{3}-6u=-10,u=1,x_{1}+x_{2}=2+{\sqrt {6}}} 55 + 81 2 2 + 33 3 + 45 2 6 3 = 2 + 6 + 30 + 12 6 2 = 2 + 3 2 + 2 3 + 6 2 {\displaystyle {\sqrt[{3}]{55+{\frac {81}{2}}{\sqrt {2}}+33{\sqrt {3}}+{\frac {45}{2}}{\sqrt {6}}}}={\frac {2+{\sqrt {6}}+{\sqrt {30+12{\sqrt {6}}}}}{2}}={\frac {2+3{\sqrt {2}}+2{\sqrt {3}}+{\sqrt {6}}}{2}}} 参考资料Loading content...参见Loading content...Loading related searches...Wikiwand - on Seamless Wikipedia browsing. On steroids.
![{\displaystyle {\sqrt[{m}]{{\sqrt {a}}\pm {\sqrt {b}}}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/c4adc4c97338f56eecbd4897ece5f50c9e6bbc0f)
，设![{\displaystyle x_{1}={\sqrt[{m}]{{\sqrt {a}}+{\sqrt {b}}}},x_{2}={\sqrt[{m}]{{\sqrt {a}}-{\sqrt {b}}}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/31d1db32c73626ee7aa9d72202f34f46f28749c2)
[1]
![{\displaystyle {\sqrt[{3}]{{\sqrt {27}}-{\sqrt {28}}}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/1139594fa7261f0898a0b289ffedc6b924ecbc2c)
![{\displaystyle x_{1}x_{2}={\sqrt[{3}]{27-28}}=-1}](//wikimedia.org/api/rest_v1/media/math/render/svg/3bff23c6940fd58b46e426ee72a948e9bd7bc395)
![{\displaystyle {\sqrt[{3}]{{\sqrt {27}}-{\sqrt {28}}}}={\frac {{\sqrt {3}}-{\sqrt {7}}}{2}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/de7d534fb55d064732eda722812cda94e9997d32)
![{\displaystyle {\sqrt[{m}]{k_{1}\pm k_{2}{\sqrt {a}}\pm k_{3}{\sqrt {b}}+k_{4}{\sqrt {ab}}}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/b7cbb593927e0e92a99506e9637fa3bf1eecf07d)
，设![{\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}}}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/36d54566b547789495bf3b5f95e1010a6d6d3fd7)
![{\displaystyle {\sqrt[{3}]{55+{\frac {81}{2}}{\sqrt {2}}+33{\sqrt {3}}+{\frac {45}{2}}{\sqrt {6}}}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/274f3894221520d0c0dffa03cb54b4fba19d28c7)
![{\displaystyle x_{1}x_{2}=-{\sqrt[{3}]{485+198{\sqrt {6}}}}=-5-2{\sqrt {6}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/0f6e216c542f184d1cd805175863d25b80c305ed)
![{\displaystyle {\sqrt[{3}]{55+{\frac {81}{2}}{\sqrt {2}}+33{\sqrt {3}}+{\frac {45}{2}}{\sqrt {6}}}}={\frac {2+{\sqrt {6}}+{\sqrt {30+12{\sqrt {6}}}}}{2}}={\frac {2+3{\sqrt {2}}+2{\sqrt {3}}+{\sqrt {6}}}{2}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/e67c94cc8c7cf31e68d668a905b943ed258eed63)
![{\displaystyle {\sqrt[{5}]{1+{\sqrt {2}}+{\sqrt {3}}}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/ba1e11033142e4484df3c803130933ba33a10a1f)
![{\displaystyle {\sqrt {1+2{\sqrt[{5}]{2}}+{\sqrt[{5}]{4}}}}={\sqrt {(1+{\sqrt[{5}]{2}})^{2}}}=1+{\sqrt[{5}]{2}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/745be143fd6c834b5eba77f862b64f71aa2f2b1d)
![{\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}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/37f78aefc0e8c7bba6f19ffbf32d7f74b303519b)
![{\displaystyle {\sqrt[{m}]{{\sqrt {a}}\pm {\sqrt {b}}}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/c4adc4c97338f56eecbd4897ece5f50c9e6bbc0f)
![{\displaystyle x_{1}={\sqrt[{m}]{{\sqrt {a}}+{\sqrt {b}}}},x_{2}={\sqrt[{m}]{{\sqrt {a}}-{\sqrt {b}}}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/31d1db32c73626ee7aa9d72202f34f46f28749c2)
![{\displaystyle {\sqrt[{3}]{{\sqrt {27}}-{\sqrt {28}}}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/1139594fa7261f0898a0b289ffedc6b924ecbc2c)
![{\displaystyle x_{1}x_{2}={\sqrt[{3}]{27-28}}=-1}](http://wikimedia.org/api/rest_v1/media/math/render/svg/3bff23c6940fd58b46e426ee72a948e9bd7bc395)
![{\displaystyle {\sqrt[{3}]{{\sqrt {27}}-{\sqrt {28}}}}={\frac {{\sqrt {3}}-{\sqrt {7}}}{2}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/de7d534fb55d064732eda722812cda94e9997d32)
![{\displaystyle {\sqrt[{m}]{k_{1}\pm k_{2}{\sqrt {a}}\pm k_{3}{\sqrt {b}}+k_{4}{\sqrt {ab}}}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/b7cbb593927e0e92a99506e9637fa3bf1eecf07d)
![{\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}}}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/36d54566b547789495bf3b5f95e1010a6d6d3fd7)
![{\displaystyle {\sqrt[{3}]{55+{\frac {81}{2}}{\sqrt {2}}+33{\sqrt {3}}+{\frac {45}{2}}{\sqrt {6}}}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/274f3894221520d0c0dffa03cb54b4fba19d28c7)
![{\displaystyle x_{1}x_{2}=-{\sqrt[{3}]{485+198{\sqrt {6}}}}=-5-2{\sqrt {6}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/0f6e216c542f184d1cd805175863d25b80c305ed)
![{\displaystyle {\sqrt[{3}]{55+{\frac {81}{2}}{\sqrt {2}}+33{\sqrt {3}}+{\frac {45}{2}}{\sqrt {6}}}}={\frac {2+{\sqrt {6}}+{\sqrt {30+12{\sqrt {6}}}}}{2}}={\frac {2+3{\sqrt {2}}+2{\sqrt {3}}+{\sqrt {6}}}{2}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/e67c94cc8c7cf31e68d668a905b943ed258eed63)