5) b1=-32, b2=16, q=b2÷b1=16÷(-32)=-0,5
![s5 = \frac{b1( {q}^{5} - 1)}{q - 1} = \frac{ - 32( {( - 0.5)}^{5} - 1) }{ - 0.5 - 1} = \\ \frac{ - 32 \times ( - 1 \frac{1}{32} )}{ - 1.5} = \frac{ 33}{ - 1.5} = - 22](https://tex.z-dn.net/?f=s5+%3D++%5Cfrac%7Bb1%28+%7Bq%7D%5E%7B5%7D++-+1%29%7D%7Bq+-+1%7D++%3D++%5Cfrac%7B+-+32%28+%7B%28+-+0.5%29%7D%5E%7B5%7D+-+1%29+%7D%7B+-+0.5+-+1%7D++%3D++%5C%5C++%5Cfrac%7B+-+32+%5Ctimes+%28+-+1+%5Cfrac%7B1%7D%7B32%7D+%29%7D%7B+-+1.5%7D++%3D++%5Cfrac%7B+33%7D%7B+-+1.5%7D++%3D++-+22)
6) b1=1, b2=-1/2, q=b2/b1=-1/2÷1=-1/2
![s5 = \frac{1( {( - 0.5)}^{5} - 1}{ - 0.5 - 1} = \frac{ - 1 \frac{1}{32} }{ - 1.5} = \\ - \frac{33}{32} \div ( - \frac{3}{2} ) = \frac{33}{32} \times \frac{2}{3} = \frac{11}{16}](https://tex.z-dn.net/?f=s5+%3D++%5Cfrac%7B1%28+%7B%28+-+0.5%29%7D%5E%7B5%7D++-+1%7D%7B+-+0.5+-+1%7D++%3D++%5Cfrac%7B+-+1+%5Cfrac%7B1%7D%7B32%7D+%7D%7B+-+1.5%7D++%3D++%5C%5C++-++%5Cfrac%7B33%7D%7B32%7D++%5Cdiv+%28+-++%5Cfrac%7B3%7D%7B2%7D+%29+%3D++%5Cfrac%7B33%7D%7B32%7D++%5Ctimes++%5Cfrac%7B2%7D%7B3%7D++%3D++%5Cfrac%7B11%7D%7B16%7D+)
8) c1=1, q=-2
![s5 = \frac{1( {( - 2)}^{5} - 1}{ - 2 - 1} = \frac{ - 32 - 1}{ - 3} = \frac{ - 33}{ - 3} = 11](https://tex.z-dn.net/?f=s5+%3D++%5Cfrac%7B1%28+%7B%28+-+2%29%7D%5E%7B5%7D++-+1%7D%7B+-+2+-+1%7D++%3D++%5Cfrac%7B+-+32+-+1%7D%7B+-+3%7D++%3D++%5Cfrac%7B+-+33%7D%7B+-+3%7D++%3D+11)
11+4х > 2х -7
2х > -18
х > -9
5 - 4х =< 7х - 5
0 =< 11х
х >= 0
14 +2(-х+7) =< 24 - 3(х-1)
14 + 14 - 2х =< 24+3 - 3х
28 - 2х =< 27 - 3х
1 <= -х
х >= 1
|х| + 7 <= 10
|х| <= 3
-3 < х < 3
<span> п<a<3п/2 это третья четверть</span>
<span>Значит косинус будет отрицательный, а тангенс и котангенс будут положительными</span>
<span>sin^2a+cos^2a=1</span>
<span>cos^2a=1-4/25</span>
<span>cosa=
</span>
<span>тангенс равен отношению синуса к косинусу</span>
<span>tga=(-2/5)/(
)</span>
<span>tga=
</span>
ctga=![\frac{\sqrt{21}}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B21%7D%7D%7B2%7D)
1)2-2сos²x-5cosx+1=0
2cos²x+5cosx-3=0
cosx=a
2a²+5a-3=0
D=25+24=49
a1=(-5-7)/4=-3⇒cosx=-3<-1 нет решения
a2=(-5+7)/4=1/2⇒cosx=1/2⇒x=+-π/6+2πn,n∈Z
2)sin2x+cos2x=0/cos2x≠0
tg2x+1=0⇒tg2x=-1⇒2x=-π/4+πn.n∈Z⇒x=-π/8+πn/2,n∈Z