\(\left|2x-5\right|=x+2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=x+2\\2x-5=-x-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-x=2+5\\2x+x=-2+5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\3x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)
Vậy ...
\(\left|2x-5\right|=x+2\)
C1:
\(\left|2x-5\right|=x+2\Leftrightarrow\left|2x-5\right|^2=\left(x+2\right)^2\\ \Leftrightarrow\left(2x-5\right)^2-\left(x+2\right)^2=0\\ \Leftrightarrow\left(2x-5+x+2\right)\left(2x-5-x-2\right)=0\\ \Leftrightarrow3\left(x-7\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)
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C2:
nếu
\(x< \dfrac{5}{2}\Rightarrow\left|2x-5\right|=5-2x\)
\(x\ge\dfrac{5}{2}\Rightarrow\left|2x-5\right|=2x-5\)
từ 2 đk trên, ta có:
\(\left[{}\begin{matrix}2x-5=x+2\left(x\ge\dfrac{5}{2}\right)\\5-2x=x+2\left(x< \dfrac{5}{2}\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=7\left(nhận\:vì\:7>\dfrac{5}{2}\right)\\x=1\left(nhận\:vì\:1< \dfrac{5}{2}\right)\end{matrix}\right.\)
vậy...
