А³+12а²+47а+60=а³+12а²+48а+64-а-4=(а+4)³-(а+4)=(а+4)((а+4)²-1)=(а+4)(а+4-1)(а+4+1)=(а+4)(а+3)(а+5)
А) Четыре последовательных члена геометрической прогрессии:
b, bq, bq², bq³.
![\left \{ {{b+bq^3=13} \atop {bq+bq^2=4}} \right. \\ \left \{ {{b(1+q^3)=13} \atop {bq(1+q)=4}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Bb%2Bbq%5E3%3D13%7D+%5Catop+%7Bbq%2Bbq%5E2%3D4%7D%7D+%5Cright.++%5C%5C++%5Cleft+%5C%7B+%7B%7Bb%281%2Bq%5E3%29%3D13%7D+%5Catop+%7Bbq%281%2Bq%29%3D4%7D%7D+%5Cright.)
![\left \{ {{bq(1+q)=4}} \atop {\frac{1-q+q^2}{q}=\frac{13}{4}}} \right.](https://tex.z-dn.net/?f=%5Cleft+%5C%7B+%7B%7Bbq%281%2Bq%29%3D4%7D%7D+%5Catop+%7B%5Cfrac%7B1-q%2Bq%5E2%7D%7Bq%7D%3D%5Cfrac%7B13%7D%7B4%7D%7D%7D+%5Cright.)
![\left \{ {{b=\frac{4}{q(1+q)}}} \atop {4-4q+4q^2=13q}}} \right. \\ \left \{ {{4q^2-17q+4=0}}} \atop {b=\frac{4}{q(1+q)}}} \right.](https://tex.z-dn.net/?f=%5Cleft+%5C%7B+%7B%7Bb%3D%5Cfrac%7B4%7D%7Bq%281%2Bq%29%7D%7D%7D+%5Catop+%7B4-4q%2B4q%5E2%3D13q%7D%7D%7D+%5Cright.+%5C%5C+%5Cleft+%5C%7B+%7B%7B4q%5E2-17q%2B4%3D0%7D%7D%7D+%5Catop+%7Bb%3D%5Cfrac%7B4%7D%7Bq%281%2Bq%29%7D%7D%7D+%5Cright.)
4q² - 17q + 4 = 0
D = 289 - 64 = 225
q = 1/4 или 4
Если q = 1/4, тогда
![b_{1}= \frac{4}{ \frac{1}{4}*(1+\frac{1}{4})}= \frac{64}{5},b_{2}=\frac{16}{5},b_{3}=\frac{4}{5},b_{4}=\frac{1}{5}](https://tex.z-dn.net/?f=b_%7B1%7D%3D+%5Cfrac%7B4%7D%7B+%5Cfrac%7B1%7D%7B4%7D%2A%281%2B%5Cfrac%7B1%7D%7B4%7D%29%7D%3D+%5Cfrac%7B64%7D%7B5%7D%2Cb_%7B2%7D%3D%5Cfrac%7B16%7D%7B5%7D%2Cb_%7B3%7D%3D%5Cfrac%7B4%7D%7B5%7D%2Cb_%7B4%7D%3D%5Cfrac%7B1%7D%7B5%7D)
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Если q = 4, тогда
![b_{1}= \frac{4}{4*(1+4)}=\frac{1}{5}, b_{2}=\frac{4}{5}, b_{3}=\frac{16}{5}, b_{4}=\frac{64}{5}](https://tex.z-dn.net/?f=b_%7B1%7D%3D+%5Cfrac%7B4%7D%7B4%2A%281%2B4%29%7D%3D%5Cfrac%7B1%7D%7B5%7D%2C++b_%7B2%7D%3D%5Cfrac%7B4%7D%7B5%7D%2C+b_%7B3%7D%3D%5Cfrac%7B16%7D%7B5%7D%2C+b_%7B4%7D%3D%5Cfrac%7B64%7D%7B5%7D)
.
Т.е. в обоих случаях члены прогрессии: 1/5, 4/5, 16/5, 64/5.
б) Три последовательных члена геометрической прогрессии: b, bq, bq².
![\left \{ {{b+bq+bq^2=19} \atop {b^2+b^2q^2+b^2q^4=133}} \right. \\ \left \{ {{b(1+q+q^2)=19} \atop {b^2(1+q^2+q^4)=133}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Bb%2Bbq%2Bbq%5E2%3D19%7D+%5Catop+%7Bb%5E2%2Bb%5E2q%5E2%2Bb%5E2q%5E4%3D133%7D%7D+%5Cright.+%5C%5C+%5Cleft+%5C%7B+%7B%7Bb%281%2Bq%2Bq%5E2%29%3D19%7D+%5Catop+%7Bb%5E2%281%2Bq%5E2%2Bq%5E4%29%3D133%7D%7D+%5Cright.)
![\left \{ {{b^2(1+q+q^2)^2=361} \atop {b^2(1+q^2+q^4)=133}} \right. \\ \left \{ {{b^2(1+q^2+q^4+2q+2q^2+2q^3)=361} \atop {b^2(1+q^2+q^4)=133}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Bb%5E2%281%2Bq%2Bq%5E2%29%5E2%3D361%7D+%5Catop+%7Bb%5E2%281%2Bq%5E2%2Bq%5E4%29%3D133%7D%7D+%5Cright.+%5C%5C++%5Cleft+%5C%7B+%7B%7Bb%5E2%281%2Bq%5E2%2Bq%5E4%2B2q%2B2q%5E2%2B2q%5E3%29%3D361%7D+%5Catop+%7Bb%5E2%281%2Bq%5E2%2Bq%5E4%29%3D133%7D%7D+%5Cright.)
![\left \{ {{b(1+q+q^2)=19} \atop {2b^2q(1+q+q^2)=228}} \right. \\ \left \{ {{b(1+q+q^2)=19} \atop {bq=6}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Bb%281%2Bq%2Bq%5E2%29%3D19%7D+%5Catop+%7B2b%5E2q%281%2Bq%2Bq%5E2%29%3D228%7D%7D+%5Cright.+%5C%5C++%5Cleft+%5C%7B+%7B%7Bb%281%2Bq%2Bq%5E2%29%3D19%7D+%5Catop+%7Bbq%3D6%7D%7D+%5Cright.)
![\left \{ {{b= \frac{6}{q} } \atop {\frac{6}{q}(1+q+q^2)=19}} \right. \\ \left \{ {{b= \frac{6}{q} } \atop {6+6q+6q^2=19q}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Bb%3D+%5Cfrac%7B6%7D%7Bq%7D+%7D+%5Catop+%7B%5Cfrac%7B6%7D%7Bq%7D%281%2Bq%2Bq%5E2%29%3D19%7D%7D+%5Cright.+%5C%5C+%5Cleft+%5C%7B+%7B%7Bb%3D+%5Cfrac%7B6%7D%7Bq%7D+%7D+%5Catop+%7B6%2B6q%2B6q%5E2%3D19q%7D%7D+%5Cright.)
6q² - 13q + 6 = 0
D = 169 - 144 = 25
q = 2/3 или 3/2
Если q = 2/3, тогда
![b_{1}= \frac{6}{\frac{2}{3}}=9, b_{2}=6, b_{3}=4.](https://tex.z-dn.net/?f=b_%7B1%7D%3D+%5Cfrac%7B6%7D%7B%5Cfrac%7B2%7D%7B3%7D%7D%3D9%2C++b_%7B2%7D%3D6%2C+b_%7B3%7D%3D4.)
Если q = 3/2, тогда
![b_{1}= \frac{6}{\frac{3}{2}}=4, b_{2}=6, b_{3}=9.](https://tex.z-dn.net/?f=b_%7B1%7D%3D+%5Cfrac%7B6%7D%7B%5Cfrac%7B3%7D%7B2%7D%7D%3D4%2C++b_%7B2%7D%3D6%2C+b_%7B3%7D%3D9.)
Т.е. в обоих случаях члены прогрессии: 4, 6, 9.