Ai giúp mik vs

Huhu ai giúp vs

Ta có: \(E=1+\frac{2xy}{x^2-xy+y^2}\le1+\frac{2xy}{2xy-xy}=3\)

Dấu "=" xảy ra <=> x = y 

Vậy GTLN của E là 3 tại x = y.

Ta có : 

\(P=\frac{3x^2+3xy+3y^2-2x^2-4xy-2y^2}{x^2+xy+y^2}=\frac{3\left(x^2+xy+y^2\right)-2\left(x^2+2xy+y^2\right)}{x^2+xy+y^2}\)

\(=3-\frac{2\left(x+y\right)^2}{x^2+xy+y^2}\le3\forall x;y\) Vì \(\frac{2\left(x+y\right)^2}{x^2+xy+y^2}\ge0\forall x;y\)

P đạt MIN là 3 tại x = - y 

Từ gt ta có x^2+y^^2=xy+1

=>P=(x^2+y^2)^2-2x^2y^2-x^2y^2

=(xy+1)²-2x²y²-x²y²

=x²y²+xy+1-3x²y²=-2x²y²+xy+1

=......

\(1=x^2+y^2-xy\ge2xy-xy=xy\Rightarrow xy\le1\)

\(1=x^2+y^2-xy\ge-2xy-xy=-3xy\Rightarrow xy\ge-\dfrac{1}{3}\)

\(\Rightarrow-\dfrac{1}{3}\le xy\le1\)

\(P=\left(x^2+y^2\right)^2-2\left(xy\right)^2-\left(xy\right)^2=\left(xy+1\right)^2-3\left(xy\right)^2=-2\left(xy\right)^2+2xy+1\)

Đặt \(xy=t\in\left[-\dfrac{1}{3};1\right]\)

\(P=f\left(t\right)=-2t^2+2t+1\)

\(f'\left(t\right)=-4t+2=0\Rightarrow t=\dfrac{1}{2}\)

\(f\left(-\dfrac{1}{3}\right)=\dfrac{1}{9}\) ; \(f\left(\dfrac{1}{2}\right)=\dfrac{3}{2}\) ; \(f\left(1\right)=1\)

\(\Rightarrow P_{max}=\dfrac{3}{2}\) ; \(P_{min}=\dfrac{1}{9}\)