\(x^2+4=3x-1
\)
<=> \(x^2+4-3x+2=0\)
<=> \(x^2-3x+6=0\)
<=>\(\left(x-3\right)^2=0\)
<=> x-3=0
<=> x=3
Vậy S= {3}
ĐKXĐ: \(x\in R\)
\(\sqrt{x^2+4}=3x-2\)
=>\(\left\{{}\begin{matrix}3x-2>=0\\x^2+4=\left(3x-2\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\9x^2-12x+4-x^2-4=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\8x^2-12x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\2x^2-3x=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\x\left(2x-3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)
=>x=3/2
