1/ \(|x|=\begin{cases}x\,\,nếu\,\,x\ge 0\\-x\,\,nếu\,\,x<0\end{cases}\)
TH1: \(x\ge 0→x=2x+7\\↔-x=7\\↔x=-7(KTM)\)
TH2: \(x<0→-x=2x+7\\↔-3x=7\\↔x=-\dfrac{7}{3}(TM)\)
Vậy \(S=\{-\dfrac{7}{3}\}\)
2/ \(|3x|=\begin{cases}3x\,\,nếu\,\,3x\ge 0\,\,hay\,\,x\ge 0\\-3x\,\,nếu\,\,3x<0\,\,hay\,\,x<0\end{cases}\)
TH1: \(x\ge 0→3x=10+2x\\↔x=10(TM)\)
TH2: \(x<0→-3x=10+2x\\↔-5x=10\\↔x=-2(TM)\)
Vậy \(S=\{10;-2\}\)
3/ \(|-x|=\begin{cases}-x\,\,nếu\,\,-x\ge 0\,\,hay\,\,x\le 0\\x\,\,nếu\,\,-x<0\,\,hay\,\,x>0\end{cases}\)
TH1: \(x\le 0→-x=5x+6\\↔-6x=6\\↔x=-1(TM)\)
TH2: \(x>0→x=5x+6\\↔-4x=6\\↔x=-\dfrac{3}{2}(KTM)\)
Vậy \(S=\{-1\}\)
4/ \(|-4x|=\begin{cases}-4x\,\,nếu\,\,-4x\ge 0\,\,hay\,\,x\le 0\\4x\,\,nếu\,\,-4x<0\,\,hay\,\,x>0\end{cases}\)
TH1: \\(x\le 0→-4x=-3x+14\\↔-x=14\\↔x=-14(TM)\)
TH2: \(x>0→4x=-3x+14\\↔7x=14\\↔x=2(TM)\)
Vậy \(S=\{-14;2\}\)
5/ \(|x+1|=\begin{cases}x+1\,\,nếu\,\,x+1\ge 0\,\,hay\,\,x\ge -1\\-x-1\,\,nếu\,\,x+1<0\,\,hay\,\,x<-1\end{cases}\)
TH1: \(x\ge -1→x+1=-2x+8\\↔3x=7\\↔x=dfrac{7}{3}(TM)\)
TH2: \(x<-1→-x-1=-2x+8\\↔x=9(KTM)\)
Vậy \(S=\{\dfrac{7}{3}\}\)
6/ \(|x-3|=\begin{cases}x-3\,\,nếu\,\,x-3\ge 0\,\,hay\,\,x\ge 3\\-x+3\,\,nếu\,\,x-3<0\,\,hay\,\,x<3\end{cases}\)
TH1: \(x\ge 3→x-3=5x+1\\↔-4x=4\\↔x=-1(KTM)\)
TH2: \(x<3→-x+3=5x+1\\↔-6x=-2\\↔x=\dfrac{1}{3}(TM)\)
Vậy \(S=\{\dfrac{1}{3}\}\)
