|x|=|x-1|+x-3
|x|-|x-1|-x=-3
ОДЗ:
x-(x-1)-x=-3,x≥0?x-1≥0
-x-(x-1)-x=-3,x<0,x-1≥0<br>x-(-(x-1))-x=-3,x≥0/x-1<0<br>-x-(-(x-1))-x=-3,x<0.x-1<0<br>а теперь просто решаем
x=4,x≥0,x≥1
x=4/3,x<0,x≥1<br>x=-2,x≥0,x≥1
x=2,x<0,x<1<br>.......................................
x=4,x∈[1;+∞)
x=4/3,x∈∅
x=-2,x∈[;1)
x=2,x∈(-∞;0)
..............................................
x=4
x∈∅
x∈∅
x∈∅
........................
x=4