Ta có: 2x-|x+3|=1
⇒|x+3|=2x-1
\(\Rightarrow\left[{}\begin{matrix}x+3=2x-1\\x+3=1-2x\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x+3-2x+1=0\\x+3-1+2x=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}4-x=0\\2+3x=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=4\\3x=2\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=4\left(tmđk\right)\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy: x=4
\(2x-\left|x+3\right|=1\)
\(\Leftrightarrow\left|x+3\right|=2x-1.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=2x-1\\x+3=1-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2x=-1-3\\x+2x=1-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=-4\\3x=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\left(-2\right):3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\frac{2}{3}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{4;-\frac{2}{3}\right\}.\)
Chúc bạn học tốt!
Ta có: 2x-|x+3|=1
⇒|x+3|=2x-1
⇒\(\left[{}\begin{matrix}x+3=2x-1\\x+3=1-2x\end{matrix}\right.\)⇒\(\left[{}\begin{matrix}x+3-2x+1=0\\x+3-1+2x=0\end{matrix}\right.\)⇒\(\left[{}\begin{matrix}-x+4=0\\3x+2=0\end{matrix}\right.\)⇒\(\left[{}\begin{matrix}-x=-4\\3x=-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=-\frac{2}{3}\end{matrix}\right.\)
Vậy: x=4
