Question

[OH-YEAH^^]

Tìm x nguyên: \(x-\dfrac{1}{y}-\dfrac{4}{xy}=-1\\\) (1 coin cho ng trl đầu, đúng)

Answer 1

[Trần Tuấn Hoàng]

-Sửa đề: x,y nguyên.

\(x-\dfrac{1}{y}-\dfrac{4}{xy}=-1\left(x\ne0;y\ne0;x\ne-1\right)\)

\(\Rightarrow x-\dfrac{1}{y}-\dfrac{4}{xy}+1=0\)

\(\Rightarrow\dfrac{x^2y}{xy}-\dfrac{x}{xy}-\dfrac{4}{xy}+\dfrac{xy}{xy}=0\)

\(\Rightarrow x^2y-x-4+xy=0\)

\(\Rightarrow xy\left(x+1\right)=x+4\)

\(\Rightarrow y=\dfrac{x+4}{x\left(x+1\right)}\)

-Vì x,y nguyên: 

\(\Rightarrow\left(x+4\right)⋮\left[x\left(x+1\right)\right]\)

\(\Rightarrow\left(x+4\right)⋮x\) và \(\left(x+4\right)⋮\left(x+1\right)\)

\(\Rightarrow4⋮x\) và \(\left(x+1+3\right)⋮\left(x+1\right)\)

\(\Rightarrow x\in\left\{1;-1;2;-2;4;-4\right\}\) và \(3⋮\left(x+1\right)\)

\(\Rightarrow x\in\left\{1;-1;2;-2;4;-4\right\}\) và \(x+1\in\left\{1;-1;3;-3\right\}\)

\(\Rightarrow x\in\left\{1;-1;2;-2;4;-4\right\}\) và \(x\in\left\{0;-2;2;-4\right\}\)

\(\Rightarrow x\in\left\{2;-2;-4\right\}\)

*\(x=2\Rightarrow y=\dfrac{2+4}{2.\left(2+1\right)}=1\)

\(x=-2\Rightarrow y=\dfrac{-2+4}{-2.\left(-2+1\right)}=1\)

\(x=-4\Rightarrow y=\dfrac{-4+4}{-4.\left(-4+1\right)}=0\left(loại\right)\)

-Vậy các cặp số (x,y) là: \(\left(2,1\right);\left(-2,1\right)\)