Question

Tìm số nguyên x,y biết xy-x+2y=3

Answer 1

Ta có:  xy - x + 2y = 3

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

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

=> x + 2; y - 1 \(\in\)Ư(1) = {1; -1}

Lập bảng:

x + 2	1	-1
y - 1	1	-1
x	-1	-3
y	2	0

Vậy ....

Answer 2

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

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

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

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

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

\(\RightarrowƯ\left(1\right)=\left\{-1;1\right\}\)

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

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

Vậy \(\left(x;y\right)=\left(-3;2\right)\)

                    \(=\left(-1;0\right)\)

Answer 3

Sorry, phần cuối để cùng dấu nhé!

Answer 4

\(\text{xy - x + 2y-3 = 0}\)

\(\text{xy - x + 2y-3+1 = 1}\)

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

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

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

\(y=\left\{2;0\right\},x=\left\{-1;-3\right\}\)

Answer 5

Thank các bạn nhé