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(x-2)²≤16/9 wiadomo, ze x²≤a² (a>0) wtedy -a≤x≤a, toz
-√16/9≤x-2≤√16/9
-4/3≤x-2≤4/3
2-4/3≤x≤2+4/3
6/3-4/3≤x≤6/3+4/3
2/3≤x≤10/3
Odpowiedz 2/3≤x≤3 1/3
2) x²>2
wiadomo, ze x²>a² (a>0) wtedy x∈(-∞;-√a)∨(√a;+∞)
toz x∈(-∞;-√2)∨(√2;+∞)
Odpowiedz x∈(-∞;-√2)∨(√2;+∞)