Воот помоему так, здесь ненадо степеней
1)(sin²4x+co²4x-2sin4xcos4x)/(sin4x-cos4x)=(sin4x-cos4x)²/(sin4x-cos4x)=sin4x-cos4x
2)(2sin²3a-sin²3a-cos²3a+2sin3acos3a)/(sin²6a+cos²6a-2sin6acos6a)= (sin²3a-cos²3a+sin6a)/(sin6a-cos6a)²=(-cos6a+sin6a)/(sin6a-cos6a)²=1/(sin6a-cos6a)
(2-x)(x+3x²)≥0
x(2-x)(1+3x)=0
x=0 x=2 x=-1/3
+ _ + _
-----------[-1/3]--------[0]----------[2]---------------
x∈(-∞;-1/3] U [0;2]
{x>0
{1-log_8x≠0⇒log_8x≠1⇒x≠8
x∈(0;8) U (8;∞)
a^3+27b^3 / a+3b=(а+3b)(a^2+3ab+b^2)/a + 3b=a^2+3ab+b^2