Пишу сразу ответ в 1) 4х³-2х
2)(х²-1)(х-4)= х³-4х²-х+4 = f ' (x)=3х²-8х-1
3) (1-х²)/х-4= ( (1-х²) ' (x-4) - (x-4) ' (1-x²) ) / (x-4)² => -2x²+8x-1+x²=> (-x²+8x-1)/ x²-8x+16
sinx>=0
2sinx*sin(П/6-x)=sqrt(3)/2
cos(2x-п/6)-cos(п/6)=sqrt(3)/2
cos(2x-п/6)=sqrt(3)>1 нет решений
sinx<0
cos(2x-п/6)=0
2x-п/6=п/2(2k+1)
x=П/12+П/4(2k+1)
sin^4(x+П/4)=(1-cos(П/2+x))^2/4=(1+sinx)^2/4
sin^4x=(1-cosx)^2/4
(1-cosx)^2+(1+sinx)^2=1
1-2cosx+cos^2x+1+2sinx+sin^2x=1
2-2cosx+2sinx=0
1-cosx+sinx=0
sin^2x/2+sinx/2cosx/2=0
sinx/2(sinx/2+cosx/2)=0
x/2=Пk x=2пk
tgx/2=-1
x=-П/2+2Пk
(2х+3у)²-3х(4/3х+4у)= 4х²+12ху+9у²-4х²-12ху= 9у²
9*(√3)²= 9*3= 27