\(2\sqrt{3x-1}=x^2-x+1\) \(\left(x\ge\dfrac{1}{3}\right)\)
\(\Leftrightarrow x^4+x^2+1-2x^3+2x^2-2x=12x-4\)
\(\Leftrightarrow x^4-2x^3+3x^2-14x+5=0\)
\(\Leftrightarrow\left(x^2-3x+1\right)\left(x^2+x+5\right)=0\)
\(\Leftrightarrow\left(x-\dfrac{3+\sqrt{5}}{2}\right)\left(x-\dfrac{3-\sqrt{5}}{2}\right)=0\)
(vì \(x^2+x+5=\left(x+\dfrac{1}{2}\right)^2+4,75>0\forall x\))
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt{5}}{2}\\x=\dfrac{3-\sqrt{5}}{2}\end{matrix}\right.\) (nhận)
đkxđ :
\(3x-1\ge0=>3x\ge1=>x\ge\dfrac{1}{3}\)
khi đó ta có
\(x^2-x+1=2\sqrt{3x-1}\)
=> \(\left(x^2-x+1\right)^2=\left(2\sqrt{3x-1}\right)^2\)
=> \(x^4-x^2+1-2x^3+2x^2-x=4\cdot\left(3x-1\right)\)
=> \(x^4-x^2+1-2x^3+2x^2-x-12x+4=0\)
=>\(x^4-2x^3+x^2-13x+5=0\)
đến đây thj mk chịu
thông cảm nha
