Question

giải phương trình sau: \(5x^2-x+5=\sqrt{x^4+x^2+1}\)

Answer 1

\(5x^2-x+5=\sqrt{x^4+x^2+1}\)

\(\Leftrightarrow5x^2-x+5=\sqrt{\left(x^2-x+1\right)\left(x^2+x+1\right)}\)

Đặt \(a=\sqrt{x^2-x+1};b=\sqrt{x^2+x+1}\left(a;b>0\right)\)

Pt tt: \(3a^2+2b^2=ab\)

\(\Leftrightarrow3a^2-ab+2b^2=0\) 

\(\Leftrightarrow3\left(a-\dfrac{b}{6}\right)^2+\dfrac{23}{12}b^2=0\)(vô nghiệm)

Vậy pt vô nghiệm