(3sinxcosx-sinx+ sin^2x- 3cosx)/cosx=0
cosxне=0
(3cosx-sinx)(1-sinx)=0
x=2пn
x=п/3+пn
-4²+3х+1=0
Д=9-4*(-4)*1=9+16=25=5²
х1=(-3+5)/-8=-2/-8=-1/4
х2=(-3-5)/-8=1
-4(х+1/4)(х-1)
4ab-bc+4a^2-ac=4a^2+4ab-bc-ac=4a(a+b)-c(a+b)=(a+b)(4a-c)
b1 = 88; q = 2; S5 = ?
Sn = b1 · (q^n - 1)/(q - 1)
n = 5
S5 = 88 · (2^5 - 1)/(2 - 1) = 88 · (32 - 1)/1 = 88 · 31 = 2728
Ответ: 2728
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