tìm x
hoặc \(3x-1=-(x+3)\)
TH1 : 3x - 1 = x + 3
3x - x = 3 + 1
2x = 4
x = 4 : 2
x = 2
TH2: 3x -1 = -(x + 3 )
3x -1 = x -3
3x - x = -3 + 1
2x = -2
x = -2 : 2
x = -1
Vậy x = 2 , x = -1
tìm x
\(\Rightarrow|x+1|=1-3x\)
\(\Rightarrow\left\{{}\begin{matrix}x+1=1-3x\\x+1=3x-1\end{matrix}\right.\)
Xét x + 1 = 1 - 3x
\(\Rightarrow x+3x=1-1\)
\(\Rightarrow4x=0\)
\(\Rightarrow x=0\div4\)
\(\Rightarrow x=0\)
Xét x + 1 = 3x -1
\(\Rightarrow x-3x=-1-1\)
\(\Rightarrow-2x=-2\)
\(\Rightarrow x=-2\div(-2)\)
\(\Rightarrow x=1\)
Vậy \(x\in\left\{0;1\right\}\)
\(|5x-3|-x=7\)
\(|5x-3|=7+x\)
\(|5x-3|=7x\)
\(\Rightarrow5x-3=7x\) hoặc \(5x-3=-7x\)
+TH1 : 5x - 3 = 7x
\(\Rightarrow\) 5x + 7x = 3
12x = 3
x = \(\frac{3}{12}\)
+TH2 : 5x - 3 = -7x
5x - 7x =3
-2x = 3
x = \(-\frac{3}{2}\)
Vậy \(x\in\left\{\frac{3}{12};\frac{-3}{2}\right\}\)