\(A=\dfrac{2\left(x^3+y^3\right)}{\left(x^4+y^2\right)\left(x^2+y^4\right)}=2.\dfrac{\left(x^3+y^3\right)}{x^4y^4+x^2y^2+x^6+y^6}\)

\(=2.\dfrac{\left(x^3+y^3\right)}{1+1+x^6+y^6}=2.\dfrac{x^3+y^3}{x^6+y^6+2x^3y^3}=2.\dfrac{x^3+y^3}{\left(x^3+y^3\right)^2}=\dfrac{2}{x^3+y^3}\left(1\right)\)

Áp dụng bất đẳng thức Cauchy ta có:

\(x^3+y^3+1\ge3\sqrt{xy.1}=3\)

\(\Rightarrow x^3+y^3\ge2\Rightarrow\dfrac{2}{x^3+y^3}\le1\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow A\le1\)

Dấu "=" xảy ra khi x=y=1.

Vậy MaxA là 1, đạt được khi x=y=1.

\(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0=>x^2+y^2\ge2xy\\\left(x+y\right)^2\ge0=>x^2+y^2\ge-2xy\end{matrix}\right.\)

Ta có:

\(\left\{{}\begin{matrix}2\left(x^2+y^2\right)+xy\ge5xy\\2\left(x^2+y^2\right)+xy\ge-3xy\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1\ge5xy\\1\ge-3xy\end{matrix}\right.\)

\(\Leftrightarrow-\dfrac{1}{3}\le xy\le\dfrac{1}{5}\)

Ta có:

P=\(2\left(x^2+y^2\right)^2-4x^2y^2+2+\left(x^2+y^2+2xy\right)\)

P= \(\dfrac{2\left(1-xy\right)^2}{4}-4\left(xy\right)^2+2+\left(\dfrac{1-xy}{2}+2xy\right)\)

=\(\dfrac{\left(xy\right)^2-2xy+1}{2}-4\left(xy\right)^2+2+\dfrac{3xy}{2}+\dfrac{1}{2}\)

Đặt t = xy => \(-\dfrac{1}{3}\le t\le\dfrac{1}{5}\)

Ta có : 

P= \(\dfrac{-7t^2}{2}+\dfrac{t}{2}+3=-\dfrac{7}{2}\left(t-\dfrac{1}{14}\right)^2+\dfrac{169}{56}\)

Ta có: \(-\dfrac{1}{3}-\dfrac{1}{14}\le t-\dfrac{1}{14}\le\dfrac{1}{5}-\dfrac{1}{14}\)

<=>\(-\dfrac{17}{42}\le t-\dfrac{1}{14}\le\dfrac{9}{70}\)

=> 0\(\le\left(t-\dfrac{1}{14}\right)^2\le\left(\dfrac{17}{42}\right)^2\)

\(\dfrac{169}{56}\ge P\ge\dfrac{169}{56}-\dfrac{7}{2}\left(\dfrac{17}{42}\right)^2\)

Max P= \(\dfrac{169}{56}\) => t = 1/14 => \(xy=\dfrac{1}{14}\rightarrow x^2+y^2=\dfrac{13}{14}\) => x,y=...

Min P=\(\dfrac{169}{56}-\dfrac{7}{6}\left(\dfrac{17}{42}\right)^2\) <=> \(t=xy=-\dfrac{1}{3}\)

<=> x=-y=\(\dfrac{1}{\sqrt{3}}\) 

Ta có: \(2\left(x^2+y^2\right)=1+xy\)

\(\Leftrightarrow x^2+y^2=\frac{1+xy}{2}\)

\(P=7\left(x^4+y^4\right)+4x^2y^2\)

\(=7x^4+7y^4+4x^2y^2\)

\(\Rightarrow P=28x^3+28y^3+16xy\)

\(\Leftrightarrow P=0\Leftrightarrow28x^3+28y^3+16xy=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\y=4\end{cases}}\)

\(\Rightarrow P_{Min}=15\) và \(Max_P=\frac{12}{33}\)

\(\left(x-1;y-1\right)=\left(a;b\right)\Rightarrow\left\{{}\begin{matrix}a;b>0\\a+b\le2\end{matrix}\right.\)

\(A=\dfrac{\left(a+1\right)^4}{b^2}+\dfrac{\left(b+1\right)^4}{a^2}\ge\dfrac{1}{2}\left[\dfrac{\left(a+1\right)^2}{b}+\dfrac{\left(b+1\right)^2}{a}\right]^2\)

\(A\ge\dfrac{1}{2}\left[\dfrac{\left(a+b+2\right)^2}{a+b}\right]^2\ge\dfrac{1}{2}\left[\dfrac{8\left(a+b\right)}{a+b}\right]^2=32\)

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

Biểu thức này không tồn tại max mà chỉ tồn tại min

\(P=\dfrac{-2\left(3x^2+2xy+y^2\right)+9x^2+6xy+y^2}{3x^2+2xy+y^2}=-2+\dfrac{\left(3x+y\right)^2}{2x^2+\left(x+y\right)^2}\ge-2\)

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