B₁+b₃<span>=15;
b</span>₂+b₄<span>=30;
n=10;
S</span>₁₀<span>=?
</span>
![S_{n}= \frac{ b_{1}( q^{n} -1) }{q-1}](https://tex.z-dn.net/?f=+S_%7Bn%7D%3D++%5Cfrac%7B+b_%7B1%7D%28+q%5E%7Bn%7D+-1%29+%7D%7Bq-1%7D+)
<span>
1)
b</span>₁+b₃ = 15 => b₁+b₁q² = 15 => b₁(1+q²) = 15
2)
b₂+b₄ = 30 => b₁q+b₁q³ = 30 => b₁(q+q³) = 30
3)
{b₁(1+q²) = 15
{b₁(q+q³) = 30
Разделим первое уравнение на второе и получим:
![\frac{1}{q}= \frac{1}{2}](https://tex.z-dn.net/?f=+%5Cfrac%7B1%7D%7Bq%7D%3D+%5Cfrac%7B1%7D%7B2%7D++)
![q=2](https://tex.z-dn.net/?f=q%3D2)
4)
b₁(1+2²)=15
b₁ · 5 =15
b₁ = 15 : 5
b₁ = 3
5)
![S_{10}= \frac{3*( 2^{10} -1)}{2-1} = \frac{3*(1024-1)}{1} =3*1023=3069](https://tex.z-dn.net/?f=+S_%7B10%7D%3D+%5Cfrac%7B3%2A%28+2%5E%7B10%7D+-1%29%7D%7B2-1%7D++%3D+%5Cfrac%7B3%2A%281024-1%29%7D%7B1%7D+%3D3%2A1023%3D3069)
S₁₀ = 3069
Y=x³+3x²-9x-2
y'=3x²+6x-9
y'=0
3x²+6x-9=3(x²+2x-3)=0
D=4+12=16
x=-3
x=1
y(-3)=(-3)³+3(-3)²-9*(-3)-2=-27+27+27-2=25
y(1)=-7
y(-8)=-250 наименьшее
y(8)=8³+3*8²-9*8-2=512+192-72-2=630 наибольшее
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