=>2x(y+7)+y+7=40

=>(y+7)(2x+1)=40

=>\(\left(2x+1;y+7\right)\in\left\{\left(1;40\right);\left(5;8\right);\left(-1;-40\right);\left(-5;-8\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(0;36\right);\left(2;1\right);\left(-1;-47\right);\left(-3;-15\right)\right\}\)

Gợi ý:
\(2xy+14x+y=33\)
\(\Rightarrow2x\left(y+7\right)+y+7=33+7\)
\(\Rightarrow\left(2x+1\right)\left(y+7\right)=40\)
\(\Rightarrow\left(2x+1;y+7\right)\inƯ\left(40\right)=\left\{\pm1;\pm2;\pm4;\pm5;\pm8;\pm10;\pm20;\pm40\right\}\)
Đến đây thì bạn làm tiếp nhé!

sao mình ko có cấp bậc?

=>2x(y+7)+y+7=40

=>(y+7)(2x+1)=40

mà x,y nguyên

nên \(\left(2x+1;y+7\right)\in\left\{\left(1;40\right);\left(5;8\right);\left(-1;-40\right);\left(-5;-8\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(0;33\right);\left(2;1\right);\left(-1;-47\right);\left(-3;-15\right)\right\}\)

\(14x-14y-x^2+2xy-y^2\\ =14\left(x-y\right)-\left(x^2-2xy+y^2\right)\\ =14\left(x-y\right)-\left(x-y\right)^2\\ =\left(x-y\right)\left(14-x+y\right)\)

2y-14x+y=33

=>3y-14x=33

=>3y=14x+33

=>y=14/3x+11

=>x chia hết cho 3 và y=14/3x+11

A= x^2 - 2x(y+7) + (y+7)^2 -(y+7)^2 + 6y^2 - 6y +72

=(x-y-7)^2 + 5(y^2 - 4y +4) +101

=(x-y-7)^2 + 5(y-2)^2 +101\(\ge\)101

\(\Rightarrow\)Min A= 101\(\Leftrightarrow\)x=9;y=2