\(\sqrt{x^2+6x+9}=\left|2x-1\right|\)
\(\Leftrightarrow\left(\sqrt{x^2+6x+9}\right)^2=\left|2x-1\right|^2\)
\(\Leftrightarrow x^2+6x+9=4x^2-4x+1\)
\(\Leftrightarrow3x^2-10x-8=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=4\end{matrix}\right.\)
Vậy phương trình có tập nghiệm : \(S=\left\{-\dfrac{2}{3};4\right\}\)
Lời giải:
\(\sqrt{x^2+6x+9}=|2x-1|\)
\(\Leftrightarrow \sqrt{(x+3)^2}=|2x-1|\)
\(\Leftrightarrow |x+3|=|2x-1|\)
\(\Rightarrow \left[\begin{matrix} x+3=2x-1\\ x+3=1-2x\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=4\\ x=\frac{-2}{3}\end{matrix}\right.\)
