=sin(2π+π/6)-cos(2π-π/6)+ctg(3π-π/4)=sinπ/6-cos(-π/6)+ctg(-π/4)=1/2-√3/2-1= -√3/2-1/2
С 2м помогу , 18-9х-2х^2 <span>≥ 0
2х^2+9x-18</span><span>≤0
D=81+144=225
x1= (-9+15)/4=1.5
x2=(-9-15)/4=-6
2(x-1.5)(x+6)</span><span>≤0
x пренадлежит (-6 : 1,5)</span>
2^(x+3) + 10*11^(x+2)<11^(x+3) + 2^(x+2) | : 11^(x+3) ≠0
(2/11)^(x+3) +10/11 < 1+1/11 *(2/11)^(x+2)
2/11 *(2/11)^(x+2)-1/11 *(2/11)^(x+2)<1-10/11
1/11 *(2/11)^(x+2)< 1/11
(2/11)^(x+2) <1; 2/11<1
x+2>0; x>-2
x=-1-наименьшее целое!