Докажите тождество: (1/x+2/y+x/y^2) : y^2-x^2/xy+2/x-y=-1/y надл ! !

(1/x+2/y+x/y^2) : y^2-x^2/xy+2/x-y=(x+y)^2/xy^2*xy/(y^2-x^2)+2/(x-y)=

=-(x+y)/(x-y)y+2/(x-y)=(y-x)/(x-y)*y=-1/y

2cos(3п/2+2x)=2sin2x=4sinx*cosx

3соs^2x=4sinx*cosx-sin^2x

разделим на сos^2x

3=4tgx-tg^2x

tgx=a

a^2-4a+3=0

d=16-12=4

a=(4+2)/2=3

a=(4-2)/2=1

tgx=3 x=arctg(3)+πn, n∈z.

tgx=1 x=π/4+πn, n∈z.

х^2-25=x^2-6x+9+2

-36=-6x

x=6