3. 16х⁴
4. 7х-6х-2х-4х=3,5
-5х=3,5
х=3,5:(-5)
х=-0,7
Cos(2x/3+π/4)=1/√2
2x/3+π/4=+-arcos(1/√2)+2πn, n∈Z
2x+π/4=π/4+2πn, n∈Z или 2x+π/4=-π/4+2πn, n∈Z
2x=2πn, n∈Z 2x==-2π/4+2πn, n∈Z
x=πn, n∈Z x=-π/2+πn, n∈Z
ответ: х1=πn, n∈Z, x2=-π/2+πn, n∈Z
= ( 4y^2 - 4xy + x^2 ) - ( 4y - 2x ) = ( 2y - x )^2 - 2( 2y - x ) = ( 2y - x )( 2y - x - 2 )