![b_{2} =b_{1}q=4](https://tex.z-dn.net/?f=+b_%7B2%7D+%3Db_%7B1%7Dq%3D4+)
![\frac{{b_{1}^{2}+b_{2}^{2}+...b_{n}^{2}}}{b_{1}+b_{2}+...b_{n}}=\frac{16}{3}](https://tex.z-dn.net/?f=+%5Cfrac%7B%7Bb_%7B1%7D%5E%7B2%7D%2Bb_%7B2%7D%5E%7B2%7D%2B...b_%7Bn%7D%5E%7B2%7D%7D%7D%7Bb_%7B1%7D%2Bb_%7B2%7D%2B...b_%7Bn%7D%7D%3D%5Cfrac%7B16%7D%7B3%7D+)
Если нужно найти сумму первых 3 членов, то ограничимся ![n=3](https://tex.z-dn.net/?f=+n%3D3+)
![n=3; b_{1}, b_{2}=b_{1}q, b_{3}=b_{1}q^{2}](https://tex.z-dn.net/?f=+n%3D3%3B+b_%7B1%7D%2C+b_%7B2%7D%3Db_%7B1%7Dq%2C+b_%7B3%7D%3Db_%7B1%7Dq%5E%7B2%7D+)
![b_{1}+b_{2}+b_{3}=b_{1}(1+q+q^{2})=b_{1}(q^{2}+q+1)](https://tex.z-dn.net/?f=+b_%7B1%7D%2Bb_%7B2%7D%2Bb_%7B3%7D%3Db_%7B1%7D%281%2Bq%2Bq%5E%7B2%7D%29%3Db_%7B1%7D%28q%5E%7B2%7D%2Bq%2B1%29+)
![b_{1}^{2}, b_{2}^{2}=b_{1}^{2}q^{2}, b_{3}^{2}=b_{1}^{2}q^{4}](https://tex.z-dn.net/?f=+b_%7B1%7D%5E%7B2%7D%2C+b_%7B2%7D%5E%7B2%7D%3Db_%7B1%7D%5E%7B2%7Dq%5E%7B2%7D%2C+b_%7B3%7D%5E%7B2%7D%3Db_%7B1%7D%5E%7B2%7Dq%5E%7B4%7D+)
![b_{1}^{2}+b_{2}^{2}+b_{3}^{2}=b_{1}^{2}+b_{1}^{2}q^{2}+b_{1}^{2}q^{4}=b_{1}^{2}(1+q^{2}+q^{4})](https://tex.z-dn.net/?f=+b_%7B1%7D%5E%7B2%7D%2Bb_%7B2%7D%5E%7B2%7D%2Bb_%7B3%7D%5E%7B2%7D%3Db_%7B1%7D%5E%7B2%7D%2Bb_%7B1%7D%5E%7B2%7Dq%5E%7B2%7D%2Bb_%7B1%7D%5E%7B2%7Dq%5E%7B4%7D%3Db_%7B1%7D%5E%7B2%7D%281%2Bq%5E%7B2%7D%2Bq%5E%7B4%7D%29+)
Представим уравнение 4 степени по-другому, выделив полный квадрат:
![q^{4}+2q^{2}+1-q^{2}=(q^{2}+1)^{2}-q^{2}=(q^{2}+q+1)(q^{2}-q+1)](https://tex.z-dn.net/?f=+q%5E%7B4%7D%2B2q%5E%7B2%7D%2B1-q%5E%7B2%7D%3D%28q%5E%7B2%7D%2B1%29%5E%7B2%7D-q%5E%7B2%7D%3D%28q%5E%7B2%7D%2Bq%2B1%29%28q%5E%7B2%7D-q%2B1%29+)
Из уравнения
выразим
:
![b_{1}=\frac{4}{q}](https://tex.z-dn.net/?f=+b_%7B1%7D%3D%5Cfrac%7B4%7D%7Bq%7D++)
Получилось уравнение:
![\frac{{b_{1}^{2}(q^{2}+q+1)(q^{2}-q+1)}}{b_{1}(q^{2}+q+1)}=\frac{16}{3}](https://tex.z-dn.net/?f=+%5Cfrac%7B%7Bb_%7B1%7D%5E%7B2%7D%28q%5E%7B2%7D%2Bq%2B1%29%28q%5E%7B2%7D-q%2B1%29%7D%7D%7Bb_%7B1%7D%28q%5E%7B2%7D%2Bq%2B1%29%7D%3D%5Cfrac%7B16%7D%7B3%7D+++)
![b_{1}\frac{{(q^{2}+q+1)(q^{2}-q+1)}}{(q^{2}+q+1)}=\frac{16}{3}](https://tex.z-dn.net/?f=+b_%7B1%7D%5Cfrac%7B%7B%28q%5E%7B2%7D%2Bq%2B1%29%28q%5E%7B2%7D-q%2B1%29%7D%7D%7B%28q%5E%7B2%7D%2Bq%2B1%29%7D%3D%5Cfrac%7B16%7D%7B3%7D+)
Подставим
:
![\frac{{4(q^{2}+q+1)(q^{2}-q+1)}}{q(q^{2}+q+1)}=\frac{16}{3}](https://tex.z-dn.net/?f=+%5Cfrac%7B%7B4%28q%5E%7B2%7D%2Bq%2B1%29%28q%5E%7B2%7D-q%2B1%29%7D%7D%7Bq%28q%5E%7B2%7D%2Bq%2B1%29%7D%3D%5Cfrac%7B16%7D%7B3%7D++)
Откуда, сократив на 4 и на
получаем:
![\frac{q^{2}-q+1}{q}=\frac{4}{3}](https://tex.z-dn.net/?f=+%5Cfrac%7Bq%5E%7B2%7D-q%2B1%7D%7Bq%7D%3D%5Cfrac%7B4%7D%7B3%7D+)
Домножим обе части на
:
![3(q^{2}-q+1)=4q](https://tex.z-dn.net/?f=+3%28q%5E%7B2%7D-q%2B1%29%3D4q+)
Раскрываем скобки:
![3q^{2}-3q+3-4q=0](https://tex.z-dn.net/?f=+3q%5E%7B2%7D-3q%2B3-4q%3D0+)
![3q^{2}-7q+3=0](https://tex.z-dn.net/?f=+3q%5E%7B2%7D-7q%2B3%3D0+)
![D=b^{2}-4ac=(-7)^{2}-4*3*3=49-36=13](https://tex.z-dn.net/?f=+D%3Db%5E%7B2%7D-4ac%3D%28-7%29%5E%7B2%7D-4%2A3%2A3%3D49-36%3D13+)
![q_{1}=\frac{7+\sqrt{13}}{6}](https://tex.z-dn.net/?f=+q_%7B1%7D%3D%5Cfrac%7B7%2B%5Csqrt%7B13%7D%7D%7B6%7D+++)
![q_{2}=\frac{7-\sqrt{13}}{6}](https://tex.z-dn.net/?f=+q_%7B2%7D%3D%5Cfrac%7B7-%5Csqrt%7B13%7D%7D%7B6%7D+++)
далее найдёте сами сумму трёх первых членов
70%-23 800руб
30%- Х руб
70х=23 800*30
х=10 200
Ответ 10 200 руб
решаем методом интервалов
нули числителя: Х-5=0
Х=5
нули знаменателя: Х+6=0
Х=-6
расставляем на числовой прямой
+ - +
――ᴼ―――ᴼ―――
-6 5
Ответ:(-6;5)
если то ,что в скобках это степени,то
a)1/125*625=5
а б вообще ничего не поняла