Question

Answer 1

ĐKXĐ : \(x\ge-\dfrac{1}{3}\)

\(x^2+x-4\sqrt{3x+1}+6=0\)

<=> \(\left(x^2-2x+1\right)+\left(3x+1-4\sqrt{3x+1}+4\right)=0\)

<=> \(\left(x-1\right)^2+\left(\sqrt{3x+1}-2\right)^2=0\)

<=> \(\left\{{}\begin{matrix}x-1=0\\\sqrt{3x+1}-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\3x+1=4\end{matrix}\right.\Leftrightarrow x=1\)

Vậy x = 1 là nghiệm phương trình