1.
\(\left|2x-1\right|=-19-x\)
\(2x-1=\pm\left(-19-x\right)\)
TH1:
\(2x-1=-19-x\)
\(2x+x=-19-1\)
\(3x=-20\)
\(x=-\frac{20}{3}\)
TH2:
\(2x-1=19+x\)
\(2x-x=19-1\)
\(x=18\)
Vậy x = -20/3 hoặc x = 18
2.
\(\left|4-3x\right|=2x-10\)
\(4-3x=\pm\left(2x-10\right)\)
TH1:
\(4-3x=2x-10\)
\(-3x-2x=-10-4\)
\(-5x=-14\)
\(x=\frac{14}{5}\)
TH2:
\(4-3x=-2x+10\)
\(-3x+2x=10-4\)
\(x=-6\)
Vậy x = 14/5 hoặc x = -6
3.
\(\left|x\right|=3+2x\)
\(x=\pm\left(3+2x\right)\)
TH1:
\(x=3+2x\)
\(x-2x=3\)
\(x=-3\)
TH2:
\(x=-3-2x\)
\(x+2x=-3\)
\(3x=-3\)
\(x=-1\)
1) | 2x-1| = -19 -x
Th1:
2x -1 = -19 -x
3x = -18
x= -6
Th2:
2x -1 = -(-19 -x)
2x -1 = 19 +x
x= 20
Vậy...
2) | 4- 3x| = 2x- 10
Th1:
4 - 3x = 2x -10
5x = 14
x= 14/5
Th2:
4- 3x = -(2x-10)
4 - 3x = -2x +10
x= -6
Vậy...
3) | x| = 3+ 2x
Th1:
3 + 2x = x
x= -3
Th2:
3 + 2x = -x
3x = -3
x= -1
Vậy...
