Sinxcos(π/3)+ sin(π/3)cosx=0
sin(x+ π/3)=0
x+ π/3 = πn, n∈Z
x=-π/3 +πn, n∈Z
F'(x) =( 0*(1 - x) - 2*(-1))/(1 - х)² = 2/(1 - х)²
f'(-1) = 2/2² = 1/2
-4(3-5х)=18-7(4-2х)
-12+20х=18-28+14х
20х-14х=18-28+12
6х=2
х=1/3
Y'=-98*x^(-2)+2
-98/x^2=-2
x^2=49
x1=-7
x2=7
методом интервалов точка максимума будет при -7
3a^2-12ab+12b^2 = 3(а^2-4аb+4b^2) = 3(a-2b)^2