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Ninel07
@Ninel07
August 2022
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hovsep50
X≠2,5 ; (2x²+x-6)/(2x-5)≥0 ⇔[(2x - 3)(x+2)]/(2x -5)≥0
1) 2x-3≥0 ; x+2≥0 ; 2x -5 >0 ⇒
x ≥ 1,5 ; x ≥ -2 ; x > 2,5 ⇔ x >2,5
2) 2x-3≥0 ; x+2≤0 ; 2x -5 <0 ⇒
x ≥ 1,5 ; x≤ -2 ; x<2,5 ⇔ x = ∅
3) 2x-3≤0 ; x+2≥0 ; 2x - 5<0 ⇒
x≤1,5 ; x≥-2 ; x<2,5 ⇔ x ∈[ -2; 1,5]
4) x≤ 1,5 ; x≤-2 ; x>2,5 ⇔ x = ∅
⇒ x ∈ [ -2 ; 1,5] U (2,5 ; ∞)
a) y(1) = √[(2+1-6)/(2-5)]= 1 верно
b) y(5) = √ 49/5 ≠√2
c) 2 ∉ [ -2 ; 1,5] U (2,5 ; ∞)
d) y(4) =√30/3 ≠ 4
e) y(3) = √16/1 = 4 ≠ √5
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Answers & Comments
1) 2x-3≥0 ; x+2≥0 ; 2x -5 >0 ⇒
x ≥ 1,5 ; x ≥ -2 ; x > 2,5 ⇔ x >2,5
2) 2x-3≥0 ; x+2≤0 ; 2x -5 <0 ⇒
x ≥ 1,5 ; x≤ -2 ; x<2,5 ⇔ x = ∅
3) 2x-3≤0 ; x+2≥0 ; 2x - 5<0 ⇒
x≤1,5 ; x≥-2 ; x<2,5 ⇔ x ∈[ -2; 1,5]
4) x≤ 1,5 ; x≤-2 ; x>2,5 ⇔ x = ∅
⇒ x ∈ [ -2 ; 1,5] U (2,5 ; ∞)
a) y(1) = √[(2+1-6)/(2-5)]= 1 верно
b) y(5) = √ 49/5 ≠√2
c) 2 ∉ [ -2 ; 1,5] U (2,5 ; ∞)
d) y(4) =√30/3 ≠ 4
e) y(3) = √16/1 = 4 ≠ √5