Giải thích các bước giải:

Ta có:

$x+\dfrac2x\in Q$

$\to (x+\dfrac2x)^2\in Q$

$\to x^2+2\cdot x\cdot \dfrac2x+(\dfrac2x)^2\in Q$

$\to x^2+4+\dfrac4{x^2}\in Q$

$\to x^2+\dfrac4{x^2}\in Q$

Vì $x^3\in Q\to \dfrac8{x^3}\in Q$

$\to x^3-\dfrac8{x^3}\in Z$

$\to (x-\dfrac2x)(x^2+x\cdot \dfrac2x+\dfrac4{x^2})\in Q$

$\to (x-\dfrac2x)(x^2+2+\dfrac4{x^2})\in Q$

Lại có: $ x^2+\dfrac4{x^2}\in Q\to  x^2+\dfrac4{x^2}+2\in Q$

$\to x-\dfrac2x\in Q$

$\to (x-\dfrac2x)+(x+\dfrac2x)\in Q$

$\to 2x\in Q$

$\to x\in Q$