pigor: ...,
IlogIx−1I+1I ≥2 ⇔ (log|x−1|+1≤−2 v log|x−1|+1≥2) i (*)
x≠1 ⇒
⇒ log|x−1|≤−3 v log|x−1|≥1 ⇒ |x−1|≤10
−3 v |x−1|≥10
1 ⇔
⇔ −10
−3≤ x−1≤ 10
−3 v x−1≤−10 v x−1≥10 ⇔
⇔ −0,999 ≤ x ≤ 1,001 v x ≤ −9 v x ≥11 , stąd i z (*) ⇔
⇔
x ≤ −9 v −0,999 ≤ x < 1 v 1< x ≤ 1,001 v x ≥11 ⇔
⇔
x∊ (−∞;−9>U
<−0,999;1)U
(1; 1,001>U
<11;+∞) . ...