(sinα+sin2α)/(1+cosα+cos2α)=tgα
1. sinα+sin2α=sinα+2sinαcosα=sinα*(1+2cosα)
2. 1+cosα+cos2α=1+cosα+2cos²α-1=cosα+2cos²α=cosα*(1+2cosα)
3. sinα*(1+2cosα)/(cosα*(1+2cosα))=sinα/cosα=tgα
4. tgα=tgα
По условию:
![S_3=21](https://tex.z-dn.net/?f=S_3%3D21)
и
![b_1^2+b_2^3+b_3^2=189](https://tex.z-dn.net/?f=b_1%5E2%2Bb_2%5E3%2Bb_3%5E2%3D189+)
. Используя формулу n-го члена геометрической прогрессии, имеем что
![\dfrac{b_1(1-q^3)}{1-q}=21](https://tex.z-dn.net/?f=+%5Cdfrac%7Bb_1%281-q%5E3%29%7D%7B1-q%7D%3D21+)
и
![b_1^2+b_1^2q^2+b_1^2q^4=189](https://tex.z-dn.net/?f=b_1%5E2%2Bb_1%5E2q%5E2%2Bb_1%5E2q%5E4%3D189)
Решая систему уравнений
![\displaystyle \left \{ {{ \dfrac{b_1(1-q^3)}{1-q}=21 } \atop {b_1^2+b_1^2q^2+b_1^2q^4=189}} \right.](https://tex.z-dn.net/?f=%5Cdisplaystyle++%5Cleft+%5C%7B+%7B%7B+%5Cdfrac%7Bb_1%281-q%5E3%29%7D%7B1-q%7D%3D21+%7D+%5Catop+%7Bb_1%5E2%2Bb_1%5E2q%5E2%2Bb_1%5E2q%5E4%3D189%7D%7D+%5Cright.+)
, имеем что
![\displaystyle \left \{ {{b_1(1+q+q^2)=21} \atop {b_1^2(1+q^2+q^4)=189}} \right. \Rightarrow~~~ \left \{ {{b_1(1+q+q^2)=21} \atop {b_1^2(q^2-q+1)(q^2+q+1)=189}} \right. \\ \\ \Rightarrow \left \{ {{b_1(1+q+q^2)=21} \atop {b_1(q^2-q+1)\cdot21=189}} \right. \Rightarrow~~~ \left \{ {{b_1(1+q+q^2)=21} \atop {b_1(q^2-q+1)=9}} \right. \\ \\ \Rightarrow ~~~~~~~\left \{ {{b_1= \dfrac{21}{1+q+q^2} } \atop { \dfrac{21(q^2-q+1)}{1+q+q^2} =9}} \right.](https://tex.z-dn.net/?f=%5Cdisplaystyle+%5Cleft+%5C%7B+%7B%7Bb_1%281%2Bq%2Bq%5E2%29%3D21%7D+%5Catop+%7Bb_1%5E2%281%2Bq%5E2%2Bq%5E4%29%3D189%7D%7D+%5Cright.++%5CRightarrow~~~+%5Cleft+%5C%7B+%7B%7Bb_1%281%2Bq%2Bq%5E2%29%3D21%7D+%5Catop+%7Bb_1%5E2%28q%5E2-q%2B1%29%28q%5E2%2Bq%2B1%29%3D189%7D%7D+%5Cright.+%5C%5C+%5C%5C+%5CRightarrow+%5Cleft+%5C%7B+%7B%7Bb_1%281%2Bq%2Bq%5E2%29%3D21%7D+%5Catop+%7Bb_1%28q%5E2-q%2B1%29%5Ccdot21%3D189%7D%7D+%5Cright.+%5CRightarrow~~~+%5Cleft+%5C%7B+%7B%7Bb_1%281%2Bq%2Bq%5E2%29%3D21%7D+%5Catop+%7Bb_1%28q%5E2-q%2B1%29%3D9%7D%7D+%5Cright.+%5C%5C+%5C%5C+%5CRightarrow+~~~~~~~%5Cleft+%5C%7B+%7B%7Bb_1%3D+%5Cdfrac%7B21%7D%7B1%2Bq%2Bq%5E2%7D+%7D+%5Catop+%7B+%5Cdfrac%7B21%28q%5E2-q%2B1%29%7D%7B1%2Bq%2Bq%5E2%7D+%3D9%7D%7D+%5Cright.++)
![7(q^2-q+1)=3(1+q+q^2)\\ 7q^2-7q+7=3+3q+3q^2\\ 4q^2-10q+4=0\\ 2q^2-5q+2=0](https://tex.z-dn.net/?f=7%28q%5E2-q%2B1%29%3D3%281%2Bq%2Bq%5E2%29%5C%5C+7q%5E2-7q%2B7%3D3%2B3q%2B3q%5E2%5C%5C+4q%5E2-10q%2B4%3D0%5C%5C+2q%5E2-5q%2B2%3D0++)
Решив как квадратное уравнение, получим
![q_1=0.5\\ q_2=2](https://tex.z-dn.net/?f=q_1%3D0.5%5C%5C+q_2%3D2+)
Тогда
0,5(4 - 2y) = y - 1,8
2 - y = y - 1,8
y + y = 2 + 1,8
2y = 3,8
y = 1,9
1). (21-9)*2=24. (Правая сторона)
2). 12*2=24. (Нижняя сторона)
3) 24+24=48 ( обе стороны )
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