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1noname
@1noname
August 2021
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помогите решить неравенства
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oganesbagoyan
Verified answer
2)
√((x -3)(2-x)) < 3+2x ;
3+2x ≤ 0 ,те ч ≤ -1 5 не имеет решения
0≤ (x-3)(2-x) < (3+2x)²;
⇔{0≤ (x-3)(2-x) ; (x-3)(2-x) < (3+2x)².
(x-2)(x 3) ≤ 0 ; -x² +5x - 6 < 9+12x +4x².
{x∈[ 2;3] ; 5x² +7x +15 >0. { {x∈[ 2;3] ; x∈ ( -∞; ∞).
x∈[ 2;3] .
3)
√((2x -1)/(x-2)) < 1 ;
0 ≤(2x -1)/(x-2) < 1²;
{ 2(x-1/2)/(x-2) ≥0 ; (2x-1)/(x-2)-1 <0.
{ x∈ (∞; 1/2] U (2; ∞) ; (x +1)/(x-2) <0.
[(x +1)/(x-2) <0.⇔(x +1)*(x-2) < 0 ]
{ x∈ (∞; 1/2] U (2; ∞) ; (x +1)*(x-2) <0.
{ x∈ (∞; 1/2] U (2; ∞) ; x ∈ (-1 ;2) .
x∈ (-1 ; 1/2] .
*************************************************
3) ??
√((2x -1)/(x-2))
> 1
;
(2x -1)/(x-2) >1²;
(2x-1)/(x-2) -1 > 0 ;.
(x+1)/(x-2) > 0 ;
(x+1)(x-2) >0;
x∈ (-∞; - 1) U (2; ∞)
.
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Answers & Comments
Verified answer
2) √((x -3)(2-x)) < 3+2x ;3+2x ≤ 0 ,те ч ≤ -1 5 не имеет решения
0≤ (x-3)(2-x) < (3+2x)²;
⇔{0≤ (x-3)(2-x) ; (x-3)(2-x) < (3+2x)².
(x-2)(x 3) ≤ 0 ; -x² +5x - 6 < 9+12x +4x².
{x∈[ 2;3] ; 5x² +7x +15 >0. { {x∈[ 2;3] ; x∈ ( -∞; ∞).
x∈[ 2;3] .
3) √((2x -1)/(x-2)) < 1 ;
0 ≤(2x -1)/(x-2) < 1²;
{ 2(x-1/2)/(x-2) ≥0 ; (2x-1)/(x-2)-1 <0.
{ x∈ (∞; 1/2] U (2; ∞) ; (x +1)/(x-2) <0. [(x +1)/(x-2) <0.⇔(x +1)*(x-2) < 0 ]
{ x∈ (∞; 1/2] U (2; ∞) ; (x +1)*(x-2) <0.
{ x∈ (∞; 1/2] U (2; ∞) ; x ∈ (-1 ;2) .
x∈ (-1 ; 1/2] .
*************************************************
3) ?? √((2x -1)/(x-2)) > 1 ;
(2x -1)/(x-2) >1²;
(2x-1)/(x-2) -1 > 0 ;.
(x+1)/(x-2) > 0 ;
(x+1)(x-2) >0;
x∈ (-∞; - 1) U (2; ∞) .