a) \(\left|x+5\right|-x+23=27\)
\(\Leftrightarrow\left|x+5\right|-x+23-27=0\)
\(\Leftrightarrow\left|x+5\right|-x+\left(23-27\right)=0\)
\(\Leftrightarrow\left|x+5\right|-x+\left(-4\right)=0\)
\(\Leftrightarrow\left|x+5\right|-x-4=0\)
\(\Leftrightarrow\left|x+5\right|-\left(x+4\right)=0\)
\(\Leftrightarrow\left|x+5\right|=x+4\)
Xét trường hợp 1:
\(x+5=x+4\)
\(\Leftrightarrow x+5-\left(x+4\right)=0\)
\(\Leftrightarrow x+5-x-4=0\)
\(\Leftrightarrow1=0\left(loai\right)\)
Xét trường hợp 2:
\(x+5=-\left(x+4\right)\)
\(\Leftrightarrow x+5-\left[-\left(x+4\right)\right]=0\)
\(\Leftrightarrow x+5+x+4=0\)
\(\Leftrightarrow2x+9=0\)
\(\Leftrightarrow2x=-9\)
\(\Leftrightarrow x=\dfrac{-9}{2}\left(thoa\right)\)
Vậy \(x=\dfrac{-9}{2}\)
