`@` `\text {Ans}`
`\downarrow`
`(x^2-2)^5 = 1`
`=> (x^2-2)^5 = 1^5`
`=> x^2-2=1`
`=> x^2 = 1+2`
`=> x^2=3`
`=> x = +-\sqrt {3}`
Vậy, `x \in +-\sqrt {3}.`
\(\Leftrightarrow x^2-2=\sqrt[5]{1}=1\\ \Leftrightarrow x^2=1+2=3\\ \Rightarrow x=\pm\sqrt{3}\)