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wolfstarrkk
@wolfstarrkk
October 2021
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Решите неравенство[tex]( \frac{2}{25 x^{2} +40x+7}+ \frac{25 x^{2} +40x+7}{2}) ^{2} \geq 4[/tex]
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sedinalana
Verified answer
(25x²+40x+7)=a
(2/a+a/2)²≥4
4/a²+2+a²/4-4≥0
4/a²-2+a²/4≥0
(2/a-a/2)²≥0
a≠0
25x²+40x+7≠0
D=1600-700=900
x≠(-40-30)/50=-1,2
x≠(-40+30)/50=-0,2
x∈(-∞;-1,2) U (-1,2;-0,2) U (-0,2;∞)
1 votes
Thanks 1
wolfstarrkk
Почему получился такой ответ? Решение неравенства ОДЗ?
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Answers & Comments
Verified answer
(25x²+40x+7)=a(2/a+a/2)²≥4
4/a²+2+a²/4-4≥0
4/a²-2+a²/4≥0
(2/a-a/2)²≥0
a≠0
25x²+40x+7≠0
D=1600-700=900
x≠(-40-30)/50=-1,2
x≠(-40+30)/50=-0,2
x∈(-∞;-1,2) U (-1,2;-0,2) U (-0,2;∞)