x−3∣≥1.8
x-3 \geq 1.8x−3≥1.8 или x-3 \leq -1.8x−3≤−1.8
x \geq 1.8+3x≥1.8+3 или x \leq -1.8+3x≤−1.8+3
x \geq 4.8x≥4.8 или x \leq 1.2x≤1.2
[1.2][4.8]
xx ∈ (-(− ∞ ;1.2];1.2] ∪ [4.8;+[4.8;+ ∞ ))
2)
|2-x|\ \textgreater \ \frac{1}{3}∣2−x∣ \textgreater 31
2-x\ \textgreater \ \frac{1}{3}2−x \textgreater 31 или 2-x\ \textless \ - \frac{1}{3}2−x \textless −31
-x\ \textgreater \ \frac{1}{3}-2−x \textgreater 31−2 или -x\ \textless \ - \frac{1}{3} -2−x \textless −31−2
x\ \textless \ 1 \frac{2}{3}x \textless 132 или x\ \textgreater \ 2 \frac{1}{3}x \textgreater 231
(1 2/3)(2 1/3)
xx ∈ (-(− ∞ ;1\frac{2}{3});132) ∪ (2\frac{2}{3};+(232;+ ∞ ))
3)
| 3-x|\ \textless \ 1.2∣3−x∣ \textless 1.2
\left \{ {{3-x\ \textless \ 1.2} \atop {3-x\ \textgreater \ -1.2}} \right.{3−x \textgreater −1.23−x \textless 1.2
\left \{ {{-x\ \textless \ 1.2-3} \atop {-x\ \textgreater \ -1.2-3}} \right.{−x \textgreater −1.2−3−x \textless 1.2−3
\left \{ {{-x\ \textless \ -1.8} \atop {-x\ \textgreater \ -4.2}} \right.{−x \textgreater −4.2−x \textless −1.8
\left \{ {{x\ \textgreater \ 1.8} \atop {x\ \textless \ 4.2}} \right.{x \textless 4.2x \textgreater 1.8
(1.8)(4.2)
xx ∈ (1.8;4.2)(1.8;4.2)
4)
|4+x | \leq 1.8∣4+x∣≤1.8
\left \{ {{4+x \leq 1.8} \atop { 4+x \geq -1.8}} \right.{4+x≥−1.84+x≤1.8
∈ 
∞ 
∪ 
∞ 
или 
или 
или 
∈ ∞ 
∪ 
∞ 
∪ 
f(x) = 2 (x-4)(x-6) /((5-4)(5-6))
вот именно в таком виде, догадайся почему