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desheva66
@desheva66
July 2022
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При каком tчисла 4t-7 , 4t+1 ,4t+5являются последовательными числами геометрической прогрессии?
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sangers1959
Verified answer
(4t+1)/(4t-7)=(t+5)/(4t+1)
(4t+1)²=(4t+5)*(4t-7)
16t²+8t+1=16t²-8t-35
16t=-36 |÷16
t=-2,25.
Ответ: при t=2,25.
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Answers & Comments
Verified answer
(4t+1)/(4t-7)=(t+5)/(4t+1)(4t+1)²=(4t+5)*(4t-7)
16t²+8t+1=16t²-8t-35
16t=-36 |÷16
t=-2,25.
Ответ: при t=2,25.