\(\left|2x-5\right|=x+1\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=x+1\\2x=5=-x-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-x=1+5\\2x+x=-1-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\3x=-6\Rightarrow x=-2\end{matrix}\right.\)
Vậy : \(x\in\left\{6,-2\right\}\)
|2x−5|=x+1|2x−5|=x+1
⇔[2x−5=x+12x=5=−x−1⇔[2x−5=x+12x=5=−x−1
⇔[2x−x=1+52x+x=−1−5⇔[2x−x=1+52x+x=−1−5
⇔[x=63x=−6⇒x=−2⇔[x=63x=−6⇒x=−2
Vậy : x∈{6,−2}
\(\left|2x-5\right|=x+1\)
⇔ \(\left[{}\begin{matrix}2x-5=x+1\\2x=5=-x-1\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}2x-x=1+5=6\\2x+x=\left(-1\right)-5=-6\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}1x=6\\3x=-6\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=6:1\\x=\left(-6\right):3\end{matrix}\right.\) \(\Rightarrow\) \(\left[{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{6;-2\right\}.\)
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