\(pt\Leftrightarrow\sqrt{x^2+3x}=3x-1\)
Bình phương hai vế, ta có :
\(\left\{{}\begin{matrix}3x-1\ge0\\x^2+3x=\left(3x-1\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{1}{3}\\8x^2-9x+1=0\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}x\ge\frac{1}{3}\\\left[{}\begin{matrix}x=1\\x=\frac{1}{8}\left(l\right)\end{matrix}\right.\end{matrix}\right.\Rightarrow}x=1}\)
Vậy phương trình có nghiệm x =1
\(\sqrt{x^2+3x}=3x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-1\ge0\\x^2+3x=\left(3x-1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{1}{3}\\x^2+3x=9x^2-6x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{1}{3}\\8x^2-9x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{1}{3}\\\left[{}\begin{matrix}x=1\\x=\frac{1}{8}\left(l\right)\end{matrix}\right.\end{matrix}\right.=>x=1\)
Vậy ...