Răspuns:
x ∈ (-∞ ; 1/2]
Explicație pas cu pas:
(x²+1)(2x-1) ≤ 0
x²+1 ≥ 0 , (∀) x ∈ R
2x-1 = 0 => x = 1/2
x I -∞ 1/2 +∞
x²+1 I+++++++++++++++++++++++++
2x-1 I--------------------0++++++++++++
(x²+1)(2x-1) I--------------------0++++++++++++
=> x ∈ (-∞ ; 1/2]
