Question

Giải phương trình

\(\sqrt{2.\left(4x^4+1\right)}=6x^2-10x+3\)

Answer 1

\(ĐK:x\le\frac{5-\sqrt{7}}{6},\frac{5+\sqrt{7}}{6}\le x\)

Ta có: \(8x^4+2=36x^4+9+100x^2+36x^2-60x-120x^3\)

    <=> \(28x^4-120x^3+136x^2-60x+7=0\)

    <=> \(\left(2x^2-6x+1\right)\left(14x^2-18x+7\right)=0\)

    <=> \(\orbr{\begin{cases}2x^2-6x+1=0\\14x^2-18x+7=0\end{cases}}\)

    \(TH_1:2x^2-6x+1=0\)

       <=> \(\orbr{\begin{cases}x=\frac{3+\sqrt{7}}{2}\left(n\right)\\x=\frac{3-\sqrt{7}}{2}\left(n\right)\end{cases}}\)

    \(TH_2:14x^2-18x+7=0\)

       <=> \(x\in\Phi\)( Tự c/m)

               Vậy \(S=\left\{\frac{3\pm\sqrt{7}}{2}\right\}\)