Lời giải:
Ta có `:`
`a^2=3+\sqrt[5+2\sqrt[3]]+3-\sqrt[5+2\sqrt[3]]+2\sqrt[9-(5+2\sqrt[3])]`
`=` `6+2\sqrt[4-2\sqrt[3]]`
`=` `6+2\sqrt[(\sqrt[3]-1)^2]`
`=` `6+2(\sqrt[3]-1)` `text( (vì)` `\sqrt[3]>1)`
`=` `4+2\sqrt[3]`
`=` `(\sqrt[3]+1)^2`
Vì `a>0` nên `a=\sqrt[3]+1`
`⇔` `a-1=\sqrt[3]`
`⇔` `(a-1)^2=\sqrt[3]^2`
`⇔` `a^2-2a+1=3`
`⇔` `a^2-2a+1-3=0`
`⇔` `a^2-2a-2=0` `⇒` `đpcm`
