Y`=14-14√2cosx
14√2cosx=14
cosx=1/√2
x=+-π/4+2+2πn,n∈z
x=π/4∈[0;π/2]
y(0)=-28-3,5π+0-0≈-38,5
y(π/4)=-28-3,5π+3,5π-14=--42 наим
y(π/2)=-28-3,5π+7π-0≈-17,5
{x³+y³ =28 ; x+y =4.⇔{ (x+y)³ -3xy(x+y) =28 ; x+y =4.⇔
{ x+y =4 ; xy =3. По обратной теореме Виета x и y корни уравнения :
t² -4t +3 =0 ; || t=1 корень ||
t₁=1 ; t₂=3. * * * x₁=<span>t₁=1 </span> ; y₁=<span> t₂=3 или </span> x₂=t₂=3 ; y₂ = <span>t₁=1 * * *</span>
ответ: { (1; 3) , (3;1)}.
2а^2+3a-b^3/b^2-9a^2= 2a^2-b/-3a