kutjrtehgrfedwSAQREDSWAZ

Answer.

`d,` Đặt `y=x+1/x` 

`=>y^2=(x+1/x)^2`

`=>y^2=(x+1/x).(x+1/x)`

`=>y^2=x^2+2+1/x^2`

`=>y^2-2=x^2+1/x^2`

________

Ta có: `x+1/x=x^2+1/x^2` `(x≠0)`

`=>y=y^2-2`

`=>y^2-y-2=0`

`=>y^2+y-2y-2=0`

`=>y.(y+1)-2.(y+1)=0`

`=>(y-2).(y+1)=0`

`=>`\(\left[ \begin{array}{l}y-2=0\\y+1=0\end{array} \right.\) 

`=>`\(\left[ \begin{array}{l}y=2\\y=-1\end{array} \right.\) 

Với `y=2=>x+1/x=2`

`=>x^2+1=2x`

`=>x^2-2x+1=0`

`=>x^2-x-x+1=0`

`=>x.(x-1)-(x-1)=0`

`=>(x-1).(x-1)=0`

`=>(x-1)^2=0`

`=>x-1=0`

`=>x=1`

Với `y=-1=>x+1/x=-1`

`=>x^2+1=-x`

`=>x^2+x+1=0`

`=>x^2+1/2x+1/2x+1/4+3/4=0`

`=>x.(x+1/2)+1/2 .(x+1/2)+3/4=0`

`=>(x+1/2).(x+1/2)+3/4=0`

`=>(x+1/2)^2+3/4=0` (Vô lý)

Vậy `x=1`