6xy-4x+3y=-53

=>2x(3y-2)+3y-2=-55

=>(3y-2)(2x+1)=-55

=>\(\left(2x+1\right)\left(3y-2\right)=1\cdot\left(-55\right)=\left(-1\right)\cdot55=\left(-55\right)\cdot1=55\cdot\left(-1\right)=5\cdot\left(-11\right)=\left(-11\right)\cdot5=\left(-5\right)\cdot11=11\cdot\left(-5\right)\)

=>\(\left(2x+1;3y-2\right)\in\){(1;-55);(-1;55);(-55;1);(55;-1);(5;-11);(-11;5);(-5;11);(11;-5)}

=>\(\left(x;y\right)\in\left\{\left(0;-\dfrac{53}{3}\right);\left(-1;19\right);\left(-28;1\right);\left(27;\dfrac{1}{3}\right);\left(2;-3\right);\left(-6;\dfrac{7}{3}\right);\left(-3;\dfrac{13}{3}\right);\left(5;-1\right)\right\}\)

mà x,y nguyên

nên \(\left(x;y\right)\in\left\{\left(-1;19\right);\left(-28;1\right);\left(2;-3\right);\left(5;-1\right)\right\}\)

Long không biết làm hả 

Ha ha ha cu Long ngu qua

Minh cug ngu nhu cu Lomg

Gio van

(2x+1)(3y-2)=-11.5

                     =11.-5

2x+1=5

=> x=2

3y-2=(-11)

=> y=(-3)

Con n truong hop khac thi cau tu trinh bay nhung ko ra x,y la Z

Nho tick nha

Ta có: \(6xy-8x-3y-2=0\)

\(\Leftrightarrow6xy-3y-8x+4-6=0\)

\(\Leftrightarrow3y\left(2x-1\right)-4\left(2x-1\right)=6\)

\(\Leftrightarrow\left(2x-1\right)\left(3y-4\right)=6\)

\(\Leftrightarrow\left(2x-1\right);\left(3y-4\right)\inƯ\left(6\right)\)

\(\Leftrightarrow\left(2x-1\right)\left(3y-4\right)\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

mà 2x-1 lẻ và \(2x-1\ge-1\) \(\forall x\in N\)

nên \(\left(2x-1\right)\in\left\{-1;1;-3;3\right\}\) và \(\left(3y-4\right)\in\left\{2;-2;6;-6\right\}\)

Trường hợp 1: 

\(\left\{{}\begin{matrix}2x-1=-1\\3y-4=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=0\\3y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=-\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\)

Trường hợp 2: 

\(\left\{{}\begin{matrix}2x-1=1\\3y-4=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=2\\3y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{10}{3}\left(loại\right)\end{matrix}\right.\)

Trường hợp 3: 

\(\left\{{}\begin{matrix}2x-1=-3\\3y-4=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=-2\\3y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\)

Trường hợp 4: 

\(\left\{{}\begin{matrix}2x-1=3\\3y-4=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=4\\3y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\left(nhận\right)\end{matrix}\right.\)

Vậy: (x,y)=(2;2)

pt <=> 3y(2x + 1) - 4x - 2 = -55

    <=> 3y(2x + 1) - 2(2x + 1) = -55

   <=> (3y - 2)(2x + 1) = -55 = (-1).55 = 55.(-1) = (-5).11 = 11.(-5)

                                            = (-55).1 = 1.(-55) = (-11).5 = 5.(-11)

Thay (3y - 2) và (2x + 1) vào 8 trường hợp trên là tìm được x,y 

tất nhiên phải xét xem x,y có là số nguyên hay không 

x=3,y=6,z=4

Ta có :  \(4x=2y=3z\)

\(\Rightarrow\frac{4x}{12}=\frac{2y}{12}=\frac{3z}{12}\) \(\Leftrightarrow\frac{x}{3}=\frac{y}{6}=\frac{z}{4}\)

Đặt  \(\frac{x}{3}=\frac{y}{6}=\frac{z}{4}=k\left(k\ne0\right)\)

 \(\Rightarrow\hept{\begin{cases}x=3k\\y=6k\\z=4k\end{cases}}\)

Mà  \(2x-3y+z=16\)

\(\Rightarrow2.3k-3.6k+4k=16\)

\(\Leftrightarrow6k-18k+4k=16\)

\(\Leftrightarrow k.\left(6-18+4\right)=16\)

\(\Leftrightarrow-8k=16\)

\(\Leftrightarrow k=-2\)

\(\Rightarrow\hept{\begin{cases}x=3k=-6\\y=6k=-12\\z=4k=-8\end{cases}}\)

Vậy ...

\(4x=2y=3z\Leftrightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)

Theo tính chất dãy tỉ số bằng nhau ta có :

\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{2x}{4}=\frac{3y}{9}=\frac{2x-3y+z}{4-9+4}=\frac{16}{-1}=-16\)

x/2=-16 => x=-16.2=-32

y/3=-16 => y=-16.3=-48

z/4=-16 => z=-16.4=-64

Vậy x=-32 ; y=-48 ; z=-64