Формула n члена геометрической прогрессии:
![b_n=b_1*q^{n-1}](https://tex.z-dn.net/?f=b_n%3Db_1%2Aq%5E%7Bn-1%7D)
тогда:
![b_2=b_1*q \\b_3=b_1*q^2 \\](https://tex.z-dn.net/?f=b_2%3Db_1%2Aq%0A%5C%5Cb_3%3Db_1%2Aq%5E2%0A%5C%5C)
составим систему и решаем ее:
![\left \{ {{b_1+b_2+b_3=70} \atop {b_1*b_2*b_3=8000}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Bb_1%2Bb_2%2Bb_3%3D70%7D+%5Catop+%7Bb_1%2Ab_2%2Ab_3%3D8000%7D%7D+%5Cright.+)
или
![\left \{ {{b_1+b_1*q+b_1*q^2=70} \atop {b_1*b_1*q*b_1*q^2=8000}} \right. \\ \left \{ {{b_1(1+q+q^2)=70} \atop {b_{1}^{3}*q^3=8000}} \right. \\ \left \{ {{b_1= \frac{70}{1+q+q^2} } \atop {b_1*q=20}} \right. \\\frac{70}{1+q+q^2}*q=20 \\ \frac{70q}{1+q+q^2} =20 \\70q=20+20q+20q^2 \\20q^2-50q+20=0 \\2q^2-5q+2=0 \\D=25-16=9=3^2 \\q_1= \frac{5+3}{4} =2 \\q_2= \frac{5-3}{4} = \frac{1}{2} =0,5 \\b_{1.1}= \frac{70}{1+2+4} = \frac{70}{7} =10 \\b_{1.2}= \frac{70}{1+0,5+0,25} = \frac{70}{1,75} =40](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Bb_1%2Bb_1%2Aq%2Bb_1%2Aq%5E2%3D70%7D+%5Catop+%7Bb_1%2Ab_1%2Aq%2Ab_1%2Aq%5E2%3D8000%7D%7D+%5Cright.+%0A%5C%5C+%5Cleft+%5C%7B+%7B%7Bb_1%281%2Bq%2Bq%5E2%29%3D70%7D+%5Catop+%7Bb_%7B1%7D%5E%7B3%7D%2Aq%5E3%3D8000%7D%7D+%5Cright.+%0A%5C%5C+%5Cleft+%5C%7B+%7B%7Bb_1%3D+%5Cfrac%7B70%7D%7B1%2Bq%2Bq%5E2%7D+%7D+%5Catop+%7Bb_1%2Aq%3D20%7D%7D+%5Cright.+%0A%5C%5C%5Cfrac%7B70%7D%7B1%2Bq%2Bq%5E2%7D%2Aq%3D20%0A%5C%5C+%5Cfrac%7B70q%7D%7B1%2Bq%2Bq%5E2%7D+%3D20%0A%5C%5C70q%3D20%2B20q%2B20q%5E2%0A%5C%5C20q%5E2-50q%2B20%3D0%0A%5C%5C2q%5E2-5q%2B2%3D0%0A%5C%5CD%3D25-16%3D9%3D3%5E2%0A%5C%5Cq_1%3D+%5Cfrac%7B5%2B3%7D%7B4%7D+%3D2%0A%5C%5Cq_2%3D+%5Cfrac%7B5-3%7D%7B4%7D+%3D+%5Cfrac%7B1%7D%7B2%7D+%3D0%2C5%0A%5C%5Cb_%7B1.1%7D%3D+%5Cfrac%7B70%7D%7B1%2B2%2B4%7D+%3D+%5Cfrac%7B70%7D%7B7%7D+%3D10%0A%5C%5Cb_%7B1.2%7D%3D+%5Cfrac%7B70%7D%7B1%2B0%2C5%2B0%2C25%7D+%3D+%5Cfrac%7B70%7D%7B1%2C75%7D+%3D40)
получим 2 варианта для b1 и q.
Ответ:
![1) q=2; \ b_1=10 \\2) q=0,5;\ b_1=40](https://tex.z-dn.net/?f=1%29+q%3D2%3B+%5C+b_1%3D10%0A%5C%5C2%29+q%3D0%2C5%3B%5C+b_1%3D40)
1) Пусть одна сторона а, другая сторона b. Найдем периметр прямоугольника. P= 2(a+b)=32.
Тогда а+b=16.
2) Найду площади двух квадратов: S=a^2 и S=b^2
3) Составим систему:
a+b=16
a^2+b^2=130
a=16-b
(16-b)^2+b^2=130
b^2-32b+256+b^2=130
2b^2-32b+126=0
b^2-16b+63=0
D=(-8)^2-63=1
b1= 8+1=9 b2=8-1=7
a=16-9=7 a=16-7=9
b=9 b=7
Ответ: 7, 9
Решение смотри в приложении,
в условии на фото у вас<span>9^х-10×3^х<=-9</span>
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S8=85/64
q=-1/2
S8=a1(q^8 -1)/(q-1)
a1=S8(q-1)/(q^8 -1)
a1=(85/64(-1/2-1))/(1/256-1)
a1=(85/64(-1.5))/(-255/256)
a1=2
Решение задания смотри на фотографии