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докажите неравенства:
а²+1≥2(3а-4)
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AlbertoЭ658
A²+1≥2(3a-4)
a²+1-6a-8⇒
a²-6a-7=0
По теореме Виета:
a₁+a₂=6
a₁·a₂=-7
a₁=-1
a₂=7
Подставляем:
1) 1+1≥2(3·(-1)-4)
2≥-14-верно.
2) 7+1≥2(3·7-4)
8≥34-неверно.
Ответ: -1
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Answers & Comments
a²+1-6a-8⇒
a²-6a-7=0
По теореме Виета:
a₁+a₂=6
a₁·a₂=-7
a₁=-1
a₂=7
Подставляем:
1) 1+1≥2(3·(-1)-4)
2≥-14-верно.
2) 7+1≥2(3·7-4)
8≥34-неверно.
Ответ: -1