ОДЗ: x²-5x+6>0
D=25-24=1
x₁=(5-1)/2=2
x₂=(5+1)/2=3
+ - +
------ 2 ----------- 3 ----------
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x∈(-∞; 2)U(3; +∞)
x² -5x+6 < (1/2)¹
x² -5x+6 -0.5 < 0
x² -5x +5.5 <0<br>D=25 - 22=3
x₁=(5-√3)/2 ≈ 1.63
x₂=(5+√3)/2≈ 3.36
+ - +
------ (5 - √3)/2 -------------- (5+√3)/2 ----------
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x∈((5-√3)/2; (5+√3)/2)
С учетом ОДЗ получаем:
x∈((5-√3)/2; 2)U(3; (5+√3)/2)