2(x+2y)+(x+2y)^2-2y^2-xy=0
(x+2y)(2-y+x+2y)=0
x+2y≠0 ОДЗ
x+y+2=0
x=-y-2
(2-y-3y-11)/(-2-y+2y)=3
-13-4y=3y-6
7y=-7
y=-1
x=-1
x-y=0
{ x - y =π/2 ; cosx - cosy =√2 ⇔ { x - y =π/2 ; - 2sin(x-y)/2*sin(x+y)/2 =√2 .
{ x - y =π/2 ; - 2sinπ/4*sin(x+y)/2 =√2 .
- 2sinπ/4*sin(x+y)/2 =√2 ;
-2*(1/√2)*sin(x+y)/2 =√2 ;
sin(x+y) = -1;
x+y = π+2π*k , k∈ Z .
{x+y = π+2π*k , k∈ Z ; x-y =π/2 ⇔ {2x =π+2π*k +π/2 ; 2y = π+2π*k -π/2.
{x =3/4π+ π*k ; y = π/4+ π*k , k ∈Z.
ответ : x =3/4π+ π*k , k ∈Z , y = π/4+ π*k , k ∈Z.
3(2-y)-2y-11=0
6-3y-2y-11=0
-5y=5
y=-1
x=2-(-1)
x=3
1)0,0625 и 2)0,1
Вот и всё.