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

Ta có: \(1+xy=x^2+y^2\ge2xy\Rightarrow xy\le1\)

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

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

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

Đặt \(a=xy\Rightarrow P=f\left(a\right)=-2a^2+2a+1\)

Xét hàm \(f\left(a\right)=-2a^2+2a+1\) trên \(\left[-\dfrac{1}{3};1\right]\)

\(-\dfrac{b}{2a}=\dfrac{1}{2}\in\left[-\dfrac{1}{3};1\right]\)

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

\(\Rightarrow M=\dfrac{3}{2}\) ; \(m=\dfrac{1}{9}\) \(\Rightarrow Mm=\dfrac{1}{6}\)

Đáp án D

Cho x,y > 0 thỏa mãn 2 ( x 2 + y 2 ) + x y = ( x + y ) ( 2 + x y ) ⇔ 2 ( x + y ) 2 - ( 2 + x y ) ( x + y ) - 3 x y = 0   (*)

Đặt x + y = u x y = v  ta đc PT bậc II: 2 u 2 - ( v + 2 ) u - 3 = 0  gải ra ta được  u = v + 2 + v 2 + 28 v + 4 4

Ta có P = 4 ( x 3 y 3 + y 3 x 3 ) - 9 ( x 2 y 2 + y 2 x 2 ) = 4 ( x y + y x ) 3 - 9 ( x y + y x ) 2 - 12 ( x y + y x ) + 18  , đặt t = ( x y + y x ) , ( t ≥ 2 ) ⇒ P = 4 t 3 - 9 t 2 - 12 t + 18  ; P ' = 6 ( 2 t 2 - 3 t + 2 ) ≥ 0  với ∀ t ≥ 2 ⇒ M i n P = P ( t 0 )  trong đó t 0 = m i n t = m i n ( x y + y x )  với x,y thỏa mãn điều kiện (*).

Ta có :

t = ( x y + y x ) = ( x + y ) 2 x y - 2 = u 2 v - 2 = ( v + 2 + v 2 + 28 v + 4 ) 2 16 v - 2 = 1 16 ( v + 2 v + v + 4 v + 28 ) 2 - 2 ≥ 1 16 ( 2 2 + 32 ) 2 - 2 = 5 2

Vậy  m i n P = P ( 5 2 ) = 4 . ( 5 2 ) 2 - 9 ( 5 2 ) 2 - 12 . 5 2 + 18 = - 23 4

\(x^2+y^2+xy=3\)

Có \(x^2+y^2\ge2xy\) \(\Rightarrow3=x^2+y^2+xy\ge2xy+xy\) \(\Leftrightarrow xy\le1\)

\(x^2+y^2\ge-2xy\) \(\Rightarrow3=x^2+y^2+xy\ge-2xy+xy\) \(\Leftrightarrow-3\le xy\) 

Đặt A= \(x^2+y^2-xy=\left(3-xy\right)-xy=3-2xy\)

mà \(-3\le xy\le1\) \(\Rightarrow9\ge3-2xy\ge1\)

=> minA=1 <=> \(\left\{{}\begin{matrix}xy=1\\x=y\end{matrix}\right.\) <=>x=y=1

maxA=9 <=>\(\left\{{}\begin{matrix}xy=-3\\x=-y\end{matrix}\right.\) <=>\(\left(x;y\right)=\left(\sqrt{3};-\sqrt{3}\right);\left(-\sqrt{3};\sqrt{3}\right)\)

Đặt \(P=x^2+y^2-xy\)

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

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

\(\dfrac{P}{3}=\dfrac{1}{3}+\dfrac{2\left(x-y\right)^2}{3\left(x^2+y^2+xy\right)}\ge\dfrac{1}{3}\Rightarrow P\ge1\)

\(P_{min}=1\) khi \(x=y=1\)

\(\dfrac{P}{3}=\dfrac{x^2+y^2-xy}{x^2+y^2+xy}=\dfrac{3\left(x^2+y^2+xy\right)-2\left(x^2+y^2+2xy\right)}{x^2+y^2+xy}=3-\dfrac{2\left(x+y\right)^2}{x^2+y^2+xy}\le3\)

\(\Rightarrow P\le9\)

\(P_{max}=9\) khi \(\left(x;y\right)=\left(\sqrt{3};-\sqrt{3}\right);\left(-\sqrt{3};\sqrt{3}\right)\)

a) Áp dụng bất đẳng thức Cosi ta có :

\(x^2+1\geq 2x\\ 4y^2+1\geq 4y\\ 9z^2+1\geq 6z\)

Suy ra \(S\leq 6\)

Dấu = xảy ra khi \(x=1;y=\frac{1}{2}; z=\frac{1}{3}\)

Đáp án C