a) Ix + 1I = x - 2
<=> x + 1 = x - 2 hay x + 1 = 2 - x
<=> x - x = -2 - 1 I <=> x + x = 2 - 1
<=> 0x = -3 (vô lí) I <=> 2x = 1
I <=> x = 1/2
b) Ix - 1I = I2xI (*)
x | 0 | 1 | |||
x - 1 | - | - | - | 0 | + |
2x | - | 0 | + | + | + |
TH1: x < 0
(*) <=> 1 - x = -2x
<=> -x + 2x = -1
<=> x = -1
TH2: 0 <= x < 1
(*) <=> 1 - x = 2x
<=> -x - 2x = -1
<=> - 3x = -1
<=> x = 1/3
TH3: x >= 1
(*) <=> x - 1 = 2x
<=> x - 2x = 1
<=> -x = 1
<=> x = -1
c) Ix - 3I + Ix - 2I = 4 (**)
x | 2 | 3 | |||
x - 2 | - | 0 | + | + | + |
x - 3 | - | - | - | 0 | + |
TH1: x < 2
(**) <=> 3 - x + 2 - x = 4
<=> -2x = 4 - 3 - 2
<=> -2x = -1
<=> x = 1/2
TH2: 2 <= x < 3
(**) <=> 3 - x + x - 2 = 4
<=> 0x = 4 + 2 + 3
<=> 0x = 9 (vô lí)
TH3: x >= 3
(**) <=> x - 3 + x - 2 = 4
<=> 2x = 4 + 2 + 3
<=> 2x = 9
<=> x = 9/2