Đáp án:
`x=0`
Giải thích các bước giải:
`sqrt(x^2+x+1)+sqrt(x^2-x+1)=sqrt(2x^2+4)`
`<=>(sqrt(x^2+x+1)+sqrt(x^2-x+1))^2=(sqrt(2x^2+4))^2`
`<=>(x^2+x+1)+(x^2-x+1)+2sqrt((sqrt(x^2+x+1))(sqrt(x^2-x+1)))=2x^2+4` `<=>2x^2+2+2sqrt((sqrt(x^2+x+1))(sqrt(x^2-x+1)))=2x^2+4`
`<=>2sqrt((sqrt(x^2+x+1))(sqrt(x^2-x+1)))=2`
`<=>sqrt((sqrt(x^2+x+1))(sqrt(x^2-x+1)))=1`
`<=>(sqrt(x^2+x+1))(sqrt(x^2-x+1))=1`
`<=>(x^2+1)^2-x^2=1`
`<=>x^4+2x^2+1-x^2=1`
`<=>x^4+x^2+1=1`
`<=>x^4+x^2=0`
`<=>x^2(x^2+1)=0`
`<=>`\(\left[\begin{array}{l}x^2=0\\x^2+1=0\end{array}\right.\)
`<=>`\(\left[\begin{array}{l}x=0\\x^2=-1(\text{loại})\end{array}\right.\)
`<=>x=0`
Vậy pt có nghiệm duy nhất `x=0`
