Tính $\sqrt[3]{9+4\sqrt{5}}+$ $\sqrt[3]{9-4\sqrt{5}}$

`\root{3}{( 9 + 4sqrt5 )}+ \root{3}{( 9 - 4sqrt5 )}`

`-` Ta đặt `a =\root{3}{( 9 + 4sqrt5 )}+ \root{3}{( 9 - 4sqrt5 )}`

`<=> a^3  = 9 + 4sqrt5 + 9 - 4sqrt 5 + 3a.\root{3}{( 9 + 4sqrt5 ).( 9 - 4sqrt5 )}`

`<=> a^3  = 18 + 3a.\root{3}{( 9^2 - (4sqrt5)^2 )}`

`<=> a^3  = 18 + 3a.\root{3}{( 81 - 80 )}`

`<=> a^3  = 18 + 3a.\root{3}{1}`

`<=> a^3  = 18 + 3a`

`<=> a^3 - 18 - 3a = 0`

`<=> a^3 + 3a^2 + 6a - 3a^2 - 9a - 18 = 0`

`<=> a^3 - 3a^2 + 3a^2 - 9a + 6a - 18 = 0`

`<=> a^2.(a - 3) + 3a.(a - 3) + 6.(a - 3) = 0`

`<=> ( a - 3).( a^2 + 3a + 6 ) = 0`

`=> {{:( a - 3 = 0),( a^2 + 3a + 6 = 0 ):}`

`<=> {{:( a = 3),(a^2 + 3a + 9/4 + 15/4 = 0 ):}`

`<=> {{:( a = 3),( ( a + 3/2)^2 + 15/4 = 0 ):}`

`-` Ta thấy `(a + 3/2)^2 >= 0` mà khi `+15/4` không thể bằng 0 cho nên phương trình này vô nghiệm.

`=> a = 3`.

Vậy `\root{3}{( 9 + 4sqrt5 )}+ \root{3}{( 9 - 4sqrt5 )} = 3`.

`N=\root{3}{9+4\sqrt{5}}+``\root{3}{9-4\sqrt{5}}`

`<=>N^3=9+4\sqrt{5}+9-4\sqrt{5}+3(``\root{3}{9+4\sqrt{5}}+``\root{3}{9-4\sqrt{5}})``\root{3}{(9+4\sqrt{5})(9-4\sqrt{5})}`

`<=>N^3=18+3N`

`<=>N^3-3N-18=0`

`<=>(N-3)(N^2+2N+6)=0`

Dễ thấy `N^2+3N+6=(A+\frac{3}{2})^2+\frac{15}{4}>0∀N`

`<=>N=3`