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DimaKanMC
@DimaKanMC
June 2021
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oganesbagoyan
Verified answer
а)
x² -2x +1 > 0 ;
(x-1)² > 0 ⇒ x ≠1 (или по другому x∈(-∞;1)U (1;∞).
б)
- 4x² +2x - 1/4 ≤ 0 ;
-( 4x² -2x + 1/4) ≤ 0 ;
( 2x -1/2)² ≥ 0⇒
x∈( -∞;∞).
в)
-x² -2x+2 <0;
x² +2x-2 >0;
x∈ (-∞; -1-√3) U(-1 +√3 ;∞) .
г)
2x²+2x -1 ≥ 0 ;
x ∈ (-∞; -(1+√3)/2)] U [(-1+√3)/2 ;∞) .
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Answers & Comments
Verified answer
а) x² -2x +1 > 0 ;(x-1)² > 0 ⇒ x ≠1 (или по другому x∈(-∞;1)U (1;∞).
б) - 4x² +2x - 1/4 ≤ 0 ;
-( 4x² -2x + 1/4) ≤ 0 ;
( 2x -1/2)² ≥ 0⇒x∈( -∞;∞).
в) -x² -2x+2 <0;
x² +2x-2 >0;
x∈ (-∞; -1-√3) U(-1 +√3 ;∞) .
г) 2x²+2x -1 ≥ 0 ;
x ∈ (-∞; -(1+√3)/2)] U [(-1+√3)/2 ;∞) .