\(\Leftrightarrow\left(x^2-4xy+4y^2\right)-\left(y^2-4y+4\right)=-1\\ \Leftrightarrow\left(x-2y\right)^2-\left(y-2\right)^2=-1\\ \Leftrightarrow\left(x-2y-y+2\right)\left(x-2y+y-2\right)=-1\\ \Leftrightarrow\left(x-3y+2\right)\left(x-y-2\right)=-1=\left(-1\right)\cdot1\)

\(TH_1:\left\{{}\begin{matrix}x-3y+2=1\\x-y-2=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-3y=-1\\x-y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\ TH_2:\left\{{}\begin{matrix}x-3y+2=-1\\x-y-2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-3y=-3\\x-y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=3\end{matrix}\right.\)

Vậy PT có nghiệm \(\left(x;y\right)\in\left\{\left(2;1\right);\left(6;3\right)\right\}\)

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

Vì \(x,y\in Z\Rightarrow\left\{{}\begin{matrix}x-y-2\in Z\\x-3y+2\in Z\\x-y-2,x-3y+2\inƯ\left(-1\right)=\left\{-1;1\right\}\end{matrix}\right.\)

Ta có bảng:

\(4xy-3x+2y=15\)

\(\Leftrightarrow x\left(4y-3\right)+2y=15\)

\(\Leftrightarrow2x\left(4y-3\right)+4y=30\)

\(\Leftrightarrow2x\left(4y-3\right)+4y-3=27\)

\(\Leftrightarrow\left(2x+1\right)\left(4x-3\right)=27\)

Ta có bảng sau: 

Vậy: \(\left(x;y\right)=\left\{\left(13;1\right);\left(-1;-6\right);\left(1;3\right);\left(-5;0\right)\right\}\)

4xy - 3x + 2y = 15

(4xy + 2y) - 3x = 15

2y(2x + 1) - 3x = 15

4y(2x + 1) - 6x = 30

4y(2x + 1) - 6x - 3 = 30 - 3

4y(2x + 1) - 3(2x + 1) = 27

(2x + 1)(4y - 3) = 27

*) TH1: 2x + 1 = -27; 4y - 3 = -1

+) 2x + 1 = -27

2x = -28

x = -14

+) 4y - 3 = -1

4y = 2

y = 1/2 (loại)

*) TH2: 2x + 1 = -9; 4y - 3 = -3

+) 2x + 1 = -9

2x = -10

x = -5

+) 4y - 3 = -3

4y = 0

y = 0

*) TH3: 2x + 1 = -1; 4y - 3 = -27

+) 2x + 1 = -1

2x = -2

x = -1

+) 4y - 3 = -27

4y = -24

y = -6

*) TH4: 2x + 1 = -3; 4y - 3 = -9

+) 2x + 1 = -3

2x = -4

x = -2

+) 4y - 3 = -9

4y = -6

y = -3/2 (loại)

*) TH5: 2x + 1 = 1; 4y - 3 = 27

+) 2x + 1 = 1

2x = 0

x = 0

+) 4y - 3 = 27

4y = 30

y = 15/2 (loại)

*) TH6: 2x + 1 = 3; 4y - 3 = 9

+) 2x + 1 = 3

2x = 2

x = 1

+) 4y - 3 = 9

4y = 12

y = 3

*) TH7: 2x + 1 = 9; 4y - 3 = 3

+) 2x + 1 = 9

2x = 8

x = 4

+) 4y - 3 = 3

4y = 6

y = 3/2 (loại)

*) TH8: 2x + 1 = 27; 4y - 3 = 1

+) 2x + 1 = 27

2x = 26

x = 13

+) 4y - 3 = 1

4y = 4

y = 1

Vậy ta tìm được các cặp giá trị (x; y) thỏa mãn:

(13; 1); (1; 3); (-1; -6); (-5; 0)

\(x^2+5y^2-4xy+10x-22y+\left|x+y+z\right|+26=0\)

\(\Leftrightarrow\left[x^2-2x\left(2y-5\right)+\left(2y-5\right)^2\right]+\left(y^2-2y+1\right)+\left|x+y+z\right|=0\)

\(\Leftrightarrow\left(x-2y+5\right)^2+\left(y-1\right)^2+\left|x+y+z\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2y+5=0\\y-1=0\\x+y+z=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=1\\z=2\end{matrix}\right.\)