Смотри.......................
2cos²x+5cosx+2=0;
cosx=t, -1≤t≤1;
2t²+5t+2=0;
D=25-16=9;
t1=(-5-3)/4=-2;
t2=(-5+3)/4=-1/2;
cosx=-1/2;
x=+-arccos(-1/2)+2πn, n∈Z;
x=+-(π-arccos1/2)+2πn, n∈Z;
x=+-(π-π/3)+2πn, n∈Z;
x=+-2π/3+2πn, n∈Z
Ответ: +-2π/3+2πn, n∈Z.
1) (150)^(3/2):(6^(3/2)=(150/6)^(3/2)=(25)^(3/2)=((5^2))^(3/2)=5^3=125
3)(2/3)^(-2)-(1/27)^(1/3)+3*(589)^0=(3/2)^2-∛(1/27)+3*1=9/4-1/3+3=5+1/4-1/3=5-1/12=4+11/12
4) (∛128+∛1/4):∛2=∛128/∛2+∛1/4/∛2=∛128/2+∛1/4/2=∛64+∛1/8=4+1/2
5) (12^(2/3)*(3^(7/3)):4^(-1/3)=(3^(2/3)*4^(2/3)*(3^(7/3)*4^(1/3)=27*4=108
3(^2/3)*3^(7/3)=3^(2/3+7/3)=3^(9/3)=3^3=27
4^(2/3)*4^(1/3)=4^(2/3+1/3)=4^(3/3)=4^1=4
12^(2/3)=(3*4)^(2/3)=3^(2/3)*4^(2/3)
1/(4^(-1/3)=4^(1/3)
(2х +1)/12x^2y + 2 -3y/18xy^2 = (2x + 1)/12xy^2 + 2 -1/6xy = =(2x + 1 + 24xy^2 - 2y)/12xy^2