\(\left|x-3\right|=9-2x\)
Th1 \(x-3=-2x+9\)
\(\Leftrightarrow3x-3=9\)
\(\Leftrightarrow3x=9+3\)
\(\Leftrightarrow3x=12\)
\(\Leftrightarrow x=4\)
Th2 : \(x-3=2x-9\)
\(\Leftrightarrow-x-3=-9\)
\(\Leftrightarrow-x=\left(-9\right)+3\)
\(\Leftrightarrow-x=-6\)
\(\Leftrightarrow x=6\) \(\text{( loại )}\)
Vậy \(x=4\)
TH1 : \(\left|x-3\right|=x-3\) khi \(x-3\ge0\Leftrightarrow x\ge3\)
TH2 : \(\left|x-3\right|=-x+3\) khi \(x-3< 0\Leftrightarrow x< 3\)
Ta có :
\(x-3=9-2x\\ \Leftrightarrow2x+x=9+3\\ \Leftrightarrow3x=12\\ \Leftrightarrow x=4\left(nhận\right)\)
Ta có :
\(-x+3=9-2x\\ \Leftrightarrow-x+2x=9-3\\ \Leftrightarrow x=6\left(loại\right)\)
Vậy S = { 4 }
