a.

\(\Leftrightarrow2x^2-4x+4y^2=4xy+4\)

\(\Leftrightarrow\left(x^2-4xy+4y^2\right)+\left(x^2-4x+4\right)=8\)

\(\Leftrightarrow\left(x-2y\right)^2+\left(x-2\right)^2=8\) (1)

Do \(\left(x-2y\right)^2\ge0;\forall x;y\)

\(\Rightarrow\left(x-2\right)^2\le8\)

\(\Rightarrow\left(x-2\right)^2=\left\{0;1;4\right\}\)

TH1: \(\left(x-2\right)^2\Rightarrow x=2\) thế vào (1)

\(\Rightarrow\left(2-2y\right)^2=8\Rightarrow\left(1-y\right)^2=2\) (ko tồn tại y nguyên t/m do 2 ko phải SCP)

TH2: \(\left(x-2\right)^2=1\Rightarrow\left(x-2y\right)^2=8-1=7\), mà 7 ko phải SCP nên pt ko có nghiệm nguyên

TH3: \(\left(x-2\right)^2=4\Rightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\) thế vào (1):

- Với \(x=0\Rightarrow\left(-2y\right)^2+4=8\Rightarrow y^2=1\Rightarrow y=\pm1\)

- Với \(x=2\Rightarrow\left(2-2y\right)^2+4=8\Rightarrow\left(1-y\right)^2=1\Rightarrow\left[{}\begin{matrix}y=0\\y=2\end{matrix}\right.\)

Vậy pt có các cặp nghiệm là: 

\(\left(x;y\right)=\left(0;1\right);\left(0;-1\right);\left(2;0\right);\left(2;2\right)\)

b.

\(\Leftrightarrow2x^2+4y^2+4xy-4x=14\)

\(\Leftrightarrow\left(x^2+4xy+4y^2\right)+\left(x^2-4x+4\right)=18\)

\(\Leftrightarrow\left(x+2y\right)^2+\left(x-2\right)^2=18\) (1)

Lý luận tương tự câu a ta được 

\(\left(x-2\right)^2\le18\Rightarrow\left(x-2\right)^2=\left\{0;1;4;9;16\right\}\)

Với \(\left(x-2\right)^2=\left\{0;1;4;16\right\}\) thì \(18-\left(x-2\right)^2\) ko phải SCP nên ko có giá trị nguyên x;y thỏa mãn

Với \(\left(x-2\right)^2=9\Rightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\) thế vào (1)

- Với \(x=5\Rightarrow\left(5+2y\right)^2+9=18\Rightarrow\left(5+2y\right)^2=9\)

\(\Rightarrow\left[{}\begin{matrix}5+2y=3\\5+2y=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}y=-1\\y=-4\end{matrix}\right.\)

- Với \(x=-1\Rightarrow\left(-1+2y\right)^2=9\Rightarrow\left[{}\begin{matrix}-1+2y=3\\-1+2y=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}y=2\\y=-1\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(5;-1\right);\left(5;-4\right);\left(-1;3\right);\left(-1;-3\right)\)

\(x^2+2y^2-4x+2y+\dfrac{9}{2}=0\)

\(x^2-4x+4+2y^2+2y+\dfrac{1}{2}=0\)

\(\left(x-2\right)^2+2\left(y+\dfrac{1}{2}\right)^2=0\)

Vì \(\left(x-2\right)^2+2\left(y+\dfrac{1}{2}\right)^2\ge0\forall x,y\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-\dfrac{1}{2}\end{matrix}\right.\)

\(x^2+2y^2-4x+2y+\dfrac{9}{2}=0\)

=>\(x^2-4x+4+2y^2+2y+\dfrac{1}{2}=0\)

=>\(\left(x-2\right)^2+2\left(y^2+y+\dfrac{1}{4}\right)=0\)

=>\(\left(x-2\right)^2+2\left(y+\dfrac{1}{2}\right)^2=0\)

mà \(\left(x-2\right)^2+2\left(y+\dfrac{1}{2}\right)^2>=0\forall x,y\)

nên \(\left\{{}\begin{matrix}x-2=0\\y+\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-\dfrac{1}{2}\end{matrix}\right.\)

A= x²+2y²-2xy-2x-2y+1015

A = x² - 2xy - 2x + y² + 2y + 1 + y² - 4y + 4 + 1010 

A = [x² - 2x(y + 1) + (y+1)² ]  + (y-2)² + 1010

A = ( x - y - 1)² + (y-2)² + 1010 \(\ge1010\forall x,y\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}x-y-1=0\\y-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)

Vậy MinA = 1010 <=> \(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\)

Bạn nên sửa lại đề là tìm GTNN

\(A=\left(x^2-2xy+y^2\right)+2\left(x-y\right)+1+y^2+4y+4+15\\ A=\left(x-y+1\right)^2+\left(y+2\right)^2+15\ge15\\ A_{min}=15\Leftrightarrow\left\{{}\begin{matrix}x=y-1\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-2\end{matrix}\right.\)

Vậy GTNN của A là 15

\(x^2+2y^2-2xy+x-2y+1=0\)

\(4x^2+8y^2-8xy+4x-8y+4=0\)

\(4x^2-4x\left(2y-1\right)+\left(2y-1\right)^2+8y^2-8y+4-\left(2y-1\right)^2=0\)

\(\left(2x-2y+1\right)^2+\left(4y^2-4y+1\right)+3=0\)

\(\left(2x-2y+1\right)^2+\left(2y-1\right)^2+3=0\) ( vô lí)

=> KL...........

vô lí

x² - 2xy + 2y² - 2x + 6y + 13 = 0 

<=> x² - 2x(y + 1) + 2y² + 6y + 13 = 0 

<=> x² - 2x(y + 1) + (y + 1)² + y² + 4y + 12 = 0 

<=> (x - y - 1)² + (y + 1)² + (y + 2)² + 8 = 0 

Vô lí do VT > 0 vs mọi x; y 

=> Ko tìm đc gtri của N

khongg lam maa ddoi co an thi an cai lon me may

\(x^2+2y^2+2xy-14y+49=0\)

\(\Leftrightarrow\left(x+y\right)^2+\left(y-7\right)^2=0\)

Dấu '=' xảy ra khi y=7 và x=-7

\(\left(x+y\right)^2+\left(y-7\right)^2=0\)
\(\Rightarrow x=-7;y=-7\)

Mong cái này giúp được bạn nhé. ☺

=>x^2-2xy+y^2+y^2+2y+1=0

=>(x-y)^2+(y+1)^2=0

=>x=y=-1

B=-2022-2023=-4045