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b) ||x + 3| - |x - 1|| = 6
Ad.a
|x| + |x-1| = x + |x-3|
-x-x+1 = x-x+3 przy opusczaniu wartości bezwzgędnej zmieniamy
znaki
-x-x-x+x = 3-1
-2x = 2 | :(-2)
x = -1
Ad.b
||x + 3| - |x - 1|| = 6
|-x-3 -x-1| =6
x+3+x+1 = 6
2x=6-3-1
2x=2 |:2
x=1