S=<span>(2a₁+d(n-1))/2)n
s10=</span>(2*5+3*9)/2)*10=185
7^x = z, 3^y = t
3z - t = 12
z*t =15 решаем подстановкой t = 3z -12
подставим во 2-е уравнение:
z(3z -12) = 15
3z² -12z -15 = 0
z² -4z -5 = 0
корни 5 и -1
z₁ = 5 z₂ = -1 t₁ = 3z -12 = 3, t₂ = 3z -12 = -15
7^x = 5 7^x = -1 3^y = 3 3^y = -15
x₁ = log₇5 ∅ y = 1 ∅
Ответ:(log₇5; 1)
(3х-1)^2+(3х-1)*(3х+1)=(3х-1)^2+9х^2-1=(3х)^2-2*3х*1+1^2+9х^2-1=9х^2-2*3х+9х^2=18х^2-2*3х=18х^2-6х.
3/6=1/2
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