Tg 3x=1/√3;
3x=atctg (1/√3)+πk, k∈Z;
3x=(π/6)+πk, k∈Z;
x=(π/18) + (π/3)k, k∈Z.
О т в е т. (π/18) + (π/3)k, k∈Z.
(3х-4у)(3х+4у) = 9x^2 - 12xy + 12xy - 16y^2 = 9x^2 - 16y^2;
Ответ: А
1)=(a-d-a-d)(a-d+a+d)=-2d*2a=-4ad
2)x(x-0.5)(x+0.5)=0 x1=0 x2=0.5 x3=-0.5
3) a^2(ab+2)-4b^2(ab+2)=(ab+2)(a^2-4b^2)=(ab+2)(a-2b)(a+2b)