Ta có : \(-\left|2,5-x\right|+1,3=0\)
=> \(-\left|2,5-x\right|=-1,3\)
=> \(\left|2,5-x\right|=1,3\)
\(\Leftrightarrow\orbr{\begin{cases}2,5-x=1,3\\2,5-x=-1,3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2,5-1,3\\x=2,5+1,3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1,2\\x=3,8\end{cases}}\)
Th1 :\(\left|2,5-x\right|=2,5-x\) khi \(2,5-x\ge0\Leftrightarrow x\le2,5\) ta có
\(-\left(2,5-x\right)+1,3=0\Leftrightarrow-2,5+x+1,3=0\)
\(\Leftrightarrow x+1,3=2,5\Leftrightarrow x=1,2\left(tm\right)\)
Th2 : \(\left|2,5-x\right|=-\left(2,5-x\right)=-2,5+x\) khi \(2,5-x< 0\Leftrightarrow x>2,5\) ta có
\(-\left(-2,5+x\right)+1,3=0\Leftrightarrow2,5-x+1,3\)
\(\Leftrightarrow-x+1,3=-2,5\Leftrightarrow-x=-3,8\Leftrightarrow x=3,8\left(tm\right)\)
vậy pt có tập nghiệm S={1,2 ; 3,8}
