\(2x^2+\dfrac{1}{x^2}+\dfrac{y^2}{4}=4\)

\(\Leftrightarrow x^2+\dfrac{1}{x^2}+x^2+\dfrac{y^2}{4}=4\left(1\right)\)

Theo Bất đẳng thức Cauchy cho các cặp số \(\left(x^2;\dfrac{1}{x^2}\right);\left(x^2;\dfrac{y^2}{4}\right)\)

\(\left\{{}\begin{matrix}x^2+\dfrac{1}{x^2}\ge2\\x^2+\dfrac{y^2}{4}\ge2.\dfrac{1}{2}xy\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2+\dfrac{1}{x^2}\ge2\\x^2+\dfrac{y^2}{4}\ge xy\end{matrix}\right.\)

Từ \(\left(1\right)\Leftrightarrow x^2+\dfrac{1}{x^2}+x^2+\dfrac{y^2}{4}\ge2+xy\)

\(\Leftrightarrow4\ge2+xy\)

\(\Leftrightarrow xy\le2\left(x;y\inℤ\right)\)

\(\Leftrightarrow Max\left(xy\right)=2\)

Dấu "=" xảy ra khi

\(xy\in\left\{-1;1;-2;2\right\}\)

\(\Leftrightarrow\left(x;y\right)\in\left\{\left(-1;-2\right);\left(1;2\right);\left(-2;-1\right);\left(2;1\right)\right\}\) thỏa mãn đề bài

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

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=\(\dfrac{1}{\sqrt{3}}\) 

a: Đặt |x-6|=a, |y+1|=b

Theo đề, ta có hệ phương trình:

\(\left\{{}\begin{matrix}2a+3b=5\\5a-4b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)

=>|x-6|=1 và |y+1|=1

\(\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{7;5\right\}\\y\in\left\{0;-2\right\}\end{matrix}\right.\)

b: Đặt |x+y|=a, |x-y|=b

Theo đề, ta có: \(\left\{{}\begin{matrix}2a-b=19\\3a+2b=17\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{55}{7}\\b=-\dfrac{23}{7}\left(loại\right)\end{matrix}\right.\)

=>HPTVN

c: Đặt |x+y|=a, |x-y|=b

Theo đề ta có: \(\left\{{}\begin{matrix}4a+3b=8\\3a-5b=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=0\end{matrix}\right.\)

=>|x+y|=2 và x=y

=>|2x|=2 và x=y

=>x=y=1 hoặc x=y=-1

Ta có: \(\left(x^2+1\right)\left(y^2+1\right)+2\left(x-y\right)\left(1-xy\right)=4\left(1+xy\right)\)

\(\Leftrightarrow x^2y^2+x^2+y^2+1-2\left(x-y\right)\left(xy-1\right)=4+4xy\)

\(\Leftrightarrow\left(x^2y^2-2xy+1\right)+\left(x^2-2xy+y^2\right)-2\left(x-y\right)\left(xy-1\right)=4\)

\(\Leftrightarrow\left(xy-1\right)^2-2\left(x-y\right)\left(xy-1\right)+\left(x-y\right)^2=4\)

\(\Leftrightarrow\left(xy-1-x+y\right)^2=4\)

\(\Leftrightarrow\left[\left(x+1\right)\left(y-1\right)\right]^2=4\)

\(\Leftrightarrow\left(x+1\right)^2\left(y-1\right)^2=4=1.4\) 

Vì \(\left(x+1\right)^2;\left(y-1\right)^2\) là các SCP và đều không âm nên ta chỉ cần xét các TH sau:

TH1: \(\hept{\begin{cases}\left(x+1\right)^2=1\\\left(y-1\right)^2=4\end{cases}}\) => \(\orbr{\begin{cases}x+1=1\\x+1=-1\end{cases}}\) và \(\orbr{\begin{cases}y-1=2\\y-1=-2\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=-2\end{cases}}\) và \(\orbr{\begin{cases}y=3\\y=-1\end{cases}}\)

TH2: \(\hept{\begin{cases}\left(x+1\right)^2=4\\\left(y-1\right)^2=1\end{cases}}\) => \(\orbr{\begin{cases}x+1=2\\x+1=-2\end{cases}}\) và \(\orbr{\begin{cases}y-1=1\\y-1=-1\end{cases}}\)

=> \(\orbr{\begin{cases}x=1\\x=-3\end{cases}}\) và \(\orbr{\begin{cases}y=2\\y=0\end{cases}}\) 

Kết luận:...

\(\left(x^2+1\right)\left(y^2+1\right)+2\left(x-y\right)\left(1-xy\right)=4\left(1+xy\right)\)

\(\Leftrightarrow\left(1-2xy+x^2y^2\right)+2\left(x-y\right)\left(1-xy\right)=4+4xy\)

\(\Leftrightarrow\left(1-xy\right)^2+2\left(x-y\right)\left(1-xy\right)+\left(x^2-2xy+y^2\right)=4\)

\(\Leftrightarrow\left(1-xy\right)^2+2\left(x-y\right)\left(1-xy\right)+\left(x-y\right)^2=4\)

\(\Leftrightarrow\left(1-xy+x-y\right)^2=4\)

\(\Leftrightarrow\left[\left(x+1\right)\left(1-y\right)\right]^2=2^2\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)\left(1-y\right)=2\\\left(x+1\right)\left(1-y\right)=-2\end{cases}}\)

Tự xét các TH