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koralinageiman
@koralinageiman
November 2021
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Подробно, пожалуйста [tex] \frac{1}{|x|-3} \leq \frac{1}{2} [/tex]
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sedinalana
Verified answer
1)x<0
1/(-x-3)-1/2≤0
1/(x+3)+1/2≥0
(2+x+3)/2(x+3)≥0
(x+5)/2(x+3)≥0
x=-5 x=-3
x≤-5 U x≥-3
x∈(-∞;-5] U (-3;0)
2)x≥0
1/(x-3)-1/2≤0
(2-x+3)/2(x-3)≤0
(5-x)/2(x-3)≤0
(x-5)/2(x-3)≥0
x=5 x=3
x<3 U x≥5
x∈(0;3) U [5;∞)
Ответx∈(-∞;-5] U (-3;0) U (0;3) U [5;∞)
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Answers & Comments
Verified answer
1)x<01/(-x-3)-1/2≤0
1/(x+3)+1/2≥0
(2+x+3)/2(x+3)≥0
(x+5)/2(x+3)≥0
x=-5 x=-3
x≤-5 U x≥-3
x∈(-∞;-5] U (-3;0)
2)x≥0
1/(x-3)-1/2≤0
(2-x+3)/2(x-3)≤0
(5-x)/2(x-3)≤0
(x-5)/2(x-3)≥0
x=5 x=3
x<3 U x≥5
x∈(0;3) U [5;∞)
Ответx∈(-∞;-5] U (-3;0) U (0;3) U [5;∞)