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Katrin2309
@Katrin2309
November 2021
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пожалуйста помогите, мне срочно
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sangers1959
Verified answer
1)
(x²-36)/(x²+6x)<0 x²+6x≠0 x* (x+6)≠0 x≠0 x≠-6
(x-6)(x+6)/(x*(x+6))<0
-∞______-______-6______+______0______-_____6______+______+∞
Ответ: x∈(-∞;-6)U(0;6).
2)
(x²+2x-8)/(16-x²)≥0 ОДЗ: 16-х²≠0 х≠+/-4
(x+4)(x-2)/((x+4)(x-4))≥0
-∞______-______-4______+______2______-______4______+_____+∞
Ответ: x∈(-4;2)U(4;+∞).
3)
(x²-6x-7)/(x²-49)≤0 ОДЗ: x²-49≠0 x≠+/-7
(x-7)(x+1)/((x-7)(x+7)≤0
-∞______+_____-7_____-_____-1______+______7______-______+∞
x∈(-7;-1]U(7;+∞).
4)
(x²-5x+4)/(5-x²)>0 ОДЗ: 5-x²≠0 x≠+/-√5
((x-4)(x-1)/((√5-x)(√5+x))>0
-∞____-___-√5___+___1____-____√5_____+_____4_____-_____+∞
Ответ: x∈(-√5;1)U(√5;4).
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Answers & Comments
Verified answer
1)(x²-36)/(x²+6x)<0 x²+6x≠0 x* (x+6)≠0 x≠0 x≠-6
(x-6)(x+6)/(x*(x+6))<0
-∞______-______-6______+______0______-_____6______+______+∞
Ответ: x∈(-∞;-6)U(0;6).
2)
(x²+2x-8)/(16-x²)≥0 ОДЗ: 16-х²≠0 х≠+/-4
(x+4)(x-2)/((x+4)(x-4))≥0
-∞______-______-4______+______2______-______4______+_____+∞
Ответ: x∈(-4;2)U(4;+∞).
3)
(x²-6x-7)/(x²-49)≤0 ОДЗ: x²-49≠0 x≠+/-7
(x-7)(x+1)/((x-7)(x+7)≤0
-∞______+_____-7_____-_____-1______+______7______-______+∞
x∈(-7;-1]U(7;+∞).
4)
(x²-5x+4)/(5-x²)>0 ОДЗ: 5-x²≠0 x≠+/-√5
((x-4)(x-1)/((√5-x)(√5+x))>0
-∞____-___-√5___+___1____-____√5_____+_____4_____-_____+∞
Ответ: x∈(-√5;1)U(√5;4).