Đáp án: `sqrt(3+sqrt5)-sqrt(3-sqrt5)-sqrt2 = 0`
Giải thích các bước giải:
Ta thấy `sqrt2 = sqrt(3+3-4)= sqrt(3+3-2.2+sqrt5-sqrt5)`
`= sqrt(3+3-2.sqrt4+sqrt5-sqrt5)= sqrt(3+3-2.sqrt(9-5)+sqrt5-sqrt5)`
`= sqrt(3+3-2.sqrt(3^2-sqrt5^2)+sqrt5-sqrt5) = sqrt(3+3-2.sqrt((3+sqrt5)(3-sqrt5))+sqrt5-sqrt5)`
`= sqrt((3+sqrt5)-2.sqrt((3+sqrt5)(3-sqrt5))+(3-sqrt5))`
`= sqrt(sqrt(3+sqrt5)^2-2.sqrt((3+sqrt5)(3-sqrt5))+sqrt(3-sqrt5)^2)`
`= sqrt(sqrt(3+sqrt5)-sqrt(3-sqrt5))^2`
`=sqrt(3+sqrt5)-sqrt(3-sqrt5)`
`=> sqrt2 = sqrt(3+sqrt5)-sqrt(3-sqrt5)`
Vậy `sqrt(3+sqrt5)-sqrt(3-sqrt5)-sqrt2`
`= sqrt(3+sqrt5)-sqrt(3-sqrt5)-(sqrt(3+sqrt5)-sqrt(3-sqrt5))`
`= sqrt(3+sqrt5)-sqrt(3-sqrt5)-sqrt(3+sqrt5)+sqrt(3-sqrt5) = 0`
