Пусть q - знаменатель прогрессии, тогда
b2 = <span>b1*</span>q
b3 = b1*<span>q²
</span> b4 = b1*<span>q³
</span>Тогда <span> b1*b2=27</span> => b1*b1*q<span>=27
</span>b3*b4=1/3 => b1*<span>q² * </span> b1*q³ = 1/3
b1² *q<span> = 27</span> => b1² = <span>27/</span>q => q > 0
b1² *q^5 = 1/3 => b1² = 1/3q^5
=> 27/q = 1/3q^5
27 * 3q^5 = q | : q
81 q^4 - 1 = 0
(9q² - 1 )(9q² + 1 ) = 0
9q² - 1 = 0
(3q - 1)(3q + 1) = 0
3q - 1 = 0 или 3q + 1 = 0
3q = 1 или 3q = - 1
<u> q = 1/3 </u> или q = - 1/3 (не удовлетворяет условию q > 0)
b1² = 27/q
b1² = 27: 1/3
b1² = 81
b1 = 9 или b1 = -9
b2 = b1*q=9*1/3 = 3 b2 = b1*q=-9*1/3 = -3
b3 = b1*q²=9*1/9 = 1 b3 = b1*q²=-9*1/9 = -1
b4 = b1*q³=9*1/27 = 1/3 b4 = b1*q³=-9*1/27 =-1/3
Ответ: <span>b1,b2,b3,b4 равны соответственно </span> 9, 3, 1 , 1/3 или
- 9, - 3, - 1 , - 1/3.
107уп-60мин
9000уп-хмин
х=9000*60:107≈5047мин или 84,1часа
(cos^2x - sin^2x)/(ctg^2x - tg^2x)=cos2x/(cos^4x-sin^4x)/(sinxcosx)^2=cos2x/(cos2x)/(sinxcosx)^2=sin^2xcos^2x