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angella73
@angella73
July 2022
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Решить неравенство
√(2х²-5х-3) > х-1
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sedinalana
Verified answer
ОДЗ
2x²-5x-3≥0
D=25+24=49
x1=(5-7)/4=-0,5
x2=(5+7)/4=3
x∈(-∞;-0,5] U [3;∞)
возведем в квадрат
2x²-5x-3>x²-2x+1
x²-3x-4>0
x1+x2=3 U x1*x2=-4
x1=-1 U x2=4
x<-1 U x>4
x∈(-∞;-1) U (4;∞)
2 votes
Thanks 3
oganesbagoyan
ответ неправильный
oganesbagoyan
Verified answer
Решить неравенство
√ (2x² - 5x
-3 ) > x - 1
---------------------------
a)
{ 2x² - 5x -3 ≥ 0
; x - 1 < 0 .
{ 2(x+1/2)(x -3) ≥ 0 ; x <1 ⇒
x ≤ 1/2
( иначе
x∈ ( -∞ ; - 1/2 ]
) .
///////////////////// [-1/2] -------------------[3 ] /////////////////////
/////////////////////////////////// (1) -----------------------------
b)
2x² - 5x
-3 > ( x - 1)
² ;
2x² - 5x -3 > x² -2x +1
;
x² - 3x - 4 >0 ;
(x-+1) (x- 4 ) >0 ⇒
x ∈ ( -∞ ; -1) U ( 4 ;∞)
.
объединяя a) и
b) получаем
:
x
∈ (-∞ ; - 1/2] U ( 4; ∞)
.
ответ : x ∈ (-∞ ; - 1/2 ] U ( 4; ∞) .
.
1 votes
Thanks 1
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Answers & Comments
Verified answer
ОДЗ2x²-5x-3≥0
D=25+24=49
x1=(5-7)/4=-0,5
x2=(5+7)/4=3
x∈(-∞;-0,5] U [3;∞)
возведем в квадрат
2x²-5x-3>x²-2x+1
x²-3x-4>0
x1+x2=3 U x1*x2=-4
x1=-1 U x2=4
x<-1 U x>4
x∈(-∞;-1) U (4;∞)
Verified answer
Решить неравенство√ (2x² - 5x -3 ) > x - 1
---------------------------
a)
{ 2x² - 5x -3 ≥ 0 ; x - 1 < 0 .
{ 2(x+1/2)(x -3) ≥ 0 ; x <1 ⇒ x ≤ 1/2 ( иначе x∈ ( -∞ ; - 1/2 ] ) .
///////////////////// [-1/2] -------------------[3 ] /////////////////////
/////////////////////////////////// (1) -----------------------------
b)
2x² - 5x -3 > ( x - 1)² ;
2x² - 5x -3 > x² -2x +1 ;
x² - 3x - 4 >0 ;
(x-+1) (x- 4 ) >0 ⇒ x ∈ ( -∞ ; -1) U ( 4 ;∞) .
объединяя a) и b) получаем : x ∈ (-∞ ; - 1/2] U ( 4; ∞) .
ответ : x ∈ (-∞ ; - 1/2 ] U ( 4; ∞) .
.