\(\left(x^2+x+1\right)^2=3\left(x^4+x^2+1\right)\)
\(\Leftrightarrow\left(x^2+x+1\right)^2=3\left[\left(x^4+2x^2+1\right)-x^2\right]\)
\(\Leftrightarrow\left(x^2+x+1\right)^2=3\left[\left(x^2+1\right)^2-x^2\right]\)
\(\Leftrightarrow\left(x^2+x+1\right)^2=3\left(x^2+x+1\right)\left(x^2-x+1\right)\)
Ta thấy \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\) nên phương trình đã cho tương đương:
\(x^2+x+1=3\left(x^2-x+1\right)\)
\(\Leftrightarrow2x^2-4x+2=0\)
\(\Leftrightarrow2\left(x-1\right)^2=0\Leftrightarrow x=1\)
Vậy phương trình đã cho có nghiệm duy nhất là x=1.
=>(x^2+x+1)^2=3[x^4+2x^2+1-x^2]
=>(x^2+x+1)^2=3(x^2+x+1)(x^2-x+1)
=>(x^2+x+1)(3x^2-3x+3-x^2-x-1)=0
=>(x^2+x+1)(2x^2-4x+2)=0
=>x=1
