\(2xy+y=13-2x\)

\(\Rightarrow2xy+y+2x=13\)

\(\Rightarrow2x\left(y+1\right)+\left(y+1\right)=14\)

\(\Rightarrow\left(y+1\right)\left(2x+1\right)=14\)

Vì \(x,y\in Z\Rightarrow\left(y+1\right)\left(2x+1\right)\in Z\Rightarrow\left(y+1\right)\left(2x+1\right)\inƯ\left(14\right)=\left\{\pm1;\pm2;\pm7;\pm14\right\}\)

Xét các trường hợp, lập bảng, kết luận.

Đến đây bạn tự làm nhé.

\(a)xy+3x-2y=11\)

\(\Leftrightarrow xy+3x-2y-6=5\)

\(\Leftrightarrow x\left(y+3\right)-2\left(y+3\right)=5\)

\(\Leftrightarrow\left(y+3\right)\left(x-2\right)=5\)

\(\Leftrightarrow\hept{\begin{cases}y+3=-1\\x-2=-5\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-4\\x=-3\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y+3=1\\x-2=5\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-2\\x=7\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y+3=-5\\x-2=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-8\\x=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y+3=5\\x-2=1\end{cases}}\Leftrightarrow\hept{\begin{cases}y=2\\x=3\end{cases}}\)

\(b)2x^2-2xy+x-y=12\)

\(\Leftrightarrow2x\left(x-y\right)+\left(x-y\right)=12\)

\(\Leftrightarrow\left(x-y\right)\left(2x+1\right)=12\)

\(\Rightarrow\left(x-y\right);\left(2x+1\right)\inƯ\left(12\right)\)

\(\RightarrowƯ\left(12\right)\in\left\{-1;1;-2;2;-3;3;-4;4;-6;6;-12;12\right\}\)

Vì 2x+1 luôn lẻ

\(\Rightarrow2x+1\in\left\{-1;1;-3;3\right\}\)

\(\Leftrightarrow\hept{\begin{cases}2x+1=-1\\x-y=-12\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=11\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2x+1=1\\x-y=12\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=-12\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2x+1=-3\\x-y=-4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2\\y=2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2x+1=3\\x-y=4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-3\end{cases}}\)

\(c)2xy-10y-x=13\)

\(\Leftrightarrow x\left(2y-1\right)-2y.5+5=18\)

\(\Leftrightarrow x\left(2y-1\right)-5\left(2y-1\right)=18\)

\(\Leftrightarrow\left(2y-1\right)\left(x-5\right)=18\)

\(\Leftrightarrow2y-1;x-5\inƯ\left(18\right)\)

\(\RightarrowƯ\left(18\right)\in\left\{-1;1;-2;2;-3;3;-6;6;-9;9;-18;18\right\}\)

Vì 2y-1  luôn lẻ

=>2y-1 thuộc {-1;1;-3;3;-9;9}

=> Làm  tương tự nhé

\(e)xy-2y^2+8y-3x=13\)

\(\Leftrightarrow xy-2y^2+2y+6y-3x-6=7\)

\(\Leftrightarrow y\left(x-2y+2\right)+3\left(-x+2y-2\right)=7\)

\(\Leftrightarrow y\left(x-2y+2\right)-3\left(x-2y+2\right)=7\)

\(\Leftrightarrow\left(x-2y+2\right)\left(y-3\right)=7\)

Tự khai triển như các câu trên.

Mình đg bận nên ko lm đc hết câu.

                                                            Bài giải

Ta có : 

\(2x+2xy-9x=13\)

\(x\left(2+2y-9\right)=13\)

\(x\left(2y-7\right)=13\)

\(\Rightarrow\text{ }\) \(x\text{ ; }\left(2y-7\right)\inƯ\left(13\right)\)

Ta có bảng : 

\(\Rightarrow\text{ }\left(x\text{ ; }y\right)=\left(-13\text{ ; }3\right),\left(13\text{ ; }10\right),\left(-1\text{ ; }3\right),\left(1\text{ ; }4\right)\)

a: =>2xy+y=7

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

=>(2x+1;y) thuộc {(1;7); (7;1); (-1;-7); (-7;-1)}

=>(x,y) thuộc {(0;7); (3;1); (-1;-7); (-4;-1)}

b: =>(2x+1)^2+(y+1)^2=179-169=10

=>((2x+1)^2;(y+1)^2) thuộc {(1;9); (9;1)}

TH1: (2x+1)^2=1 và (y+1)^2=9

=>\(\left\{{}\begin{matrix}2x+1\in\left\{1;-1\right\}\\y+1\in\left\{3;-3\right\}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{0;-1\right\}\\y\in\left\{2;-4\right\}\end{matrix}\right.\)

TH2: (2x+1)^2=9 và (y+1)^2=1

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

Các bạn làm nhanh hộ mình với ạ

a) \(\Leftrightarrow4x^2+2y^2+4xy-20x-8y+26=0\)

\(\Leftrightarrow4x^2+4x\left(y-5\right)+\left(y-5\right)^2-\left(y-5\right)^2+2y^2-8y+26=0\)

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

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

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

b) \(\Leftrightarrow\left(x^2-4x+4\right)+\left(y^2+6y+9\right)+\left(z^2-2z+1\right)=0\)

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

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

c) \(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2xz\right)+\left(x^2+2x+1\right)+\left(z^2-4z+4\right)=0\)

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

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

bạn nhấn vào  đúng 0 sẽ ra đáp án

khó quá!!!

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

\(A=x^2-2.2xy+\left(2y\right)^2+2x-4y+3\)

\(A=\left(x-2y\right)^2-2.\left(x-2y\right)+1+2\)

\(A=\left(x-2y-1\right)^2+2\ge2\)

Vậy GTNN của A=2.