Question

Tìm giá trị x,y nguyên dương trong biểu thức sau : \(\dfrac{1}{2x}+\dfrac{1}{2y}+\dfrac{1}{xy}=\dfrac{1}{2}\)

Answer 1

Lời giải:

Ta có: \(\frac{1}{2x}+\frac{1}{2y}+\frac{1}{xy}=\frac{1}{2}\)

\(\Leftrightarrow \frac{y+x}{2xy}+\frac{2}{2xy}=\frac{1}{2}\) \(\Leftrightarrow \frac{x+y+2}{2xy}=\frac{1}{2}\Leftrightarrow x+y+2=xy\)

\(\Leftrightarrow xy-x-y=2\)

\(\Leftrightarrow x(y-1)-(y-1)=3\)

\(\Leftrightarrow (x-1)(y-1)=3\)

Vì \(x,y\in\mathbb{Z}^+\Rightarrow x-1,y-1\geq 0\)

Do đó: \(\left[\begin{matrix} x-1=1,y-1=3\\ x-1=3,y-1=1\end{matrix}\right.\)

\(\Leftrightarrow \left[\begin{matrix} (x,y)=(2,4)\\ (x,y)=(4,2)\end{matrix}\right.\)

Vậy \((x,y)=(2,4)\) và hoán vị.