1
(∛(2x)+2y)(∛(4x²)-2y∛(2x)+4y²)/(∛(4x²)-2y∛(2x)+4y²) -∛(2x)=
=∛(2x)+2y-∛(2x)=2y
4
2sin(5x-π/4)=-1
sin(5x-π/4)=-1/2
5x-π/4=(-1)^(n+1)π/6+πn
5x=π/4+(-1)^(n+1)π/6+πn
x=π/20+(-1)^(n+1)π/30+πn/5,n∈z
8
-√2sinx*cosx=sinx
sinx+√2sinxcosx=0
sinx(1+√2cosx)=0
sinx=0⇒x=πn,n∈z
-5π≤πn≤-4π
-5≤n≤-4
n=-5⇒x=-5π
n=-4⇒x=-4π
1+√2cosx=0
cosx=-1/√2
x=-3π/4+2πk U x=3π/4+2πm
-5π≤-3π/4+2πk≤-4π
-20≤-3+8k≤-16
-17≤8k≤-13
-17/8≤k≤-13/8
k=-2⇒x=-3π/4-4π=-19π/4
-5π≤3π/4+2πm≤-4π
-20≤3+8m≤-16
-23≤8m≤-19
-23/8≤m≤-19/8
нет решения
Ответ x={-5π;-4π;-19π/4}