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OlegMelnikov
@OlegMelnikov
September 2021
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Решите неравенство:
[tex] |\frac{x-5}{x} |(x- x^{2} +12 ) \geq 0[/tex]
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mikael2
Verified answer
Модуль выражения всегда ≥0 по определению. x≠0
-x²+x+12≥0 x²-x-12≤0 x= 4, -3
--------- -3---------------- 4------------
+ - +
x={5}∪[-3;0)∪(0;4]
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Answers & Comments
Verified answer
Модуль выражения всегда ≥0 по определению. x≠0-x²+x+12≥0 x²-x-12≤0 x= 4, -3
--------- -3---------------- 4------------
+ - +
x={5}∪[-3;0)∪(0;4]