\(\left|2x-5\right|=\left|x+2\right|\)
\(\Rightarrow2x-5=x+2\) ; \(2x-5=-x-2\)
+) \(2x+5=x+2\)
\(\Leftrightarrow x=-3\)
+) \(2x-5=-x-2\)
\(\Leftrightarrow x=1\)
Tập nghiệm: \(S=\left\{1;-3\right\}\)
\(\left|2x-5\right|=\left|x+2\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=x+2\\2x-5=-x-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)
Vậy S=\(\left\{7;1\right\}\)
\(\Leftrightarrow\)2x-5=\(\left[{}\begin{matrix}2x-5\ge0\\2x-5>0\end{matrix}\right.\)
\(\Leftrightarrow\)x\(\ge\)\(\dfrac{5}{2}\)
\(\Leftrightarrow\)x<\(\dfrac{5}{2}\)
với x\(\ge\)\(\dfrac{5}{2}\)ta có bpt:
\(\Leftrightarrow\)2x-5=x+2
\(\Leftrightarrow\)2x-x=5+2
\(\Leftrightarrow\)x=7(nhận)
với x<\(\dfrac{5}{2}\)ta có bpt:
-(2x-5)=x+2
\(\Leftrightarrow\)-2x+5=x+2
\(\Leftrightarrow\)-2x-x=-5+2
\(\Leftrightarrow\) -3x=-3
\(\Leftrightarrow\)x=\(\dfrac{3}{3}\)=1(nhận)
vậy S=\(\left\{1;7\right\}\)
