\(0,2+\left|x-1,3\right|=1,5\)
\(=>\left|x-1,3\right|=1,5-0,2=1,3\)
=>\(\left[{}\begin{matrix}x-1,3=1,3\\x-1,3=-1,3\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=1,3+1,3=2,6\\x=-1,3+1,3=0\end{matrix}\right.\)
Vậy x ∈ { 2,6 ; 0 }
0,2+|x - 1,3|=1,5
|x - 1,3|=1,5 - 0,2
|x - 1,3|= 1,3
Th1:
x - 1,3 =1,3
x = 1,3 + 1,3
x =2,6
Th2:
x - 1,3 = -1,3
x = -1,3 + 1,3
x = 0
=> x ϵ { 2,6 ; 0 }
~hok tốt~ 
0,2+|x−1,3|=1,50,2+|x−1,3|=1,5
=>|x−1,3|=1,5−0,2=1,3=>|x−1,3|=1,5−0,2=1,3
=>[x−1,3=1,3x−1,3=−1,3[x−1,3=1,3x−1,3=−1,3
=>[x=1,3+1,3=2,6x=−1,3+1,3=0[x=1,3+1,3=2,6x=−1,3+1,3=0
Vậy x ∈ { 2,6 ; 0 }
2.
0, 2 + | x - 1, 3| = 1, 5
| x - 1, 3| = 1, 5 - 0, 2
| x - 1, 3| = 1, 3
TH1: x - 1, 3 = 1, 3
x = 1, 3 + 1, 3
x = 2, 6
TH2: x - 1, 3 = -1, 3
x = (-1, 3) + 1, 3
x = 0
Vậy x ∈ { 2, 6 ; 0}
Chúc bạn học tốt!
