`\qquad x=3+\sqrt{2}`
`\qquad y=3-\sqrt{2}`
`=>x+y=6`
`\qquad xy=(3+\sqrt{2})(3-\sqrt{2})=3^2-2=7`
Ta có:
`\qquad x^5+y^5`
`=(x^2+y^2)(x^3+y^3)-(x^2y^3+x^3y^2)`
`=[(x+y)^2-2xy][(x+y)^3-3xy(x+y)]-(xy)^2 (x+y)`
`=(6^2 -2.7).(6^3-3.7.6)-7^2 . 6`
`=22.90-294=1686`
Vậy `x^5+y^5=1686`