Question

[Tạ Ánh Dương]

x².y+x.y-x=4

Answer 1

[Duc Nguyen]

\(x^2\cdot y+x\cdot y-x=4\)

\(x\cdot y\cdot\left(x+1\right)-x=4\)

\(x\cdot y\cdot\left(x+1\right)-x-1=4-1\)

\(xy\cdot\left(x+1\right)-\left(x+1\right)=3\)

\(\left(x+1\right)\left(xy-1\right)=3\)

⇒ \((x + 1) ; (xy - 1)\) là ước của 3

⇒ \(\text{(x + 1) ; (xy - 1)}\in\left\{\pm1;\pm3\right\}\)

Ta có:

TH1:

\(\left[{}\begin{matrix}x+1=1\\xy-1=3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1-1\\xy=3+1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\xy=4\end{matrix}\right.\Rightarrow y=\dfrac{xy}{x}=\dfrac{4}{0}\) (vô nghiệm)

TH2:

\(\left[{}\begin{matrix}x+1=-1\\xy-1=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-1-1\\xy=-3+1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-2\\xy=-2\end{matrix}\right.\)\(\Rightarrow y=\dfrac{xy}{x}=\dfrac{-2}{-2}=1\) (chọn)

TH3:

\(\left[{}\begin{matrix}x+1=3\\xy-1=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3-1\\xy=1+1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\xy=2\end{matrix}\right.\)\(\Rightarrow y=\dfrac{xy}{x}=\dfrac{2}{2}=1\) (chọn)

TH4:

\(\left[{}\begin{matrix}x+1=-3\\xy-1=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-3-1\\xy=-1+1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-4\\xy=0\end{matrix}\right.\)\(\Rightarrow y=\dfrac{xy}{x}=\dfrac{0}{-4}=0\)(vô nghiệm)

Vậy (x,y)ϵ{(-2 ; 1);(2 ; 1)}