Điều kiện x, y ≠ 0

Ta có: \(\left( {2{\rm{x}} + y} \right) - \left( {2y + x} \right) = \frac{3}{{{x^2}}} - \frac{3}{{{y^2}}}\)

\( \Leftrightarrow x - y = \frac{{3\left( {{y^2} - {x^2}} \right)}}{{{x^2}{y^2}}}\)

\( \Leftrightarrow x - y = \frac{{3\left( {y - x} \right)\left( {y + x} \right)}}{{{x^2}{y^2}}}\)

\( \Leftrightarrow x - y - \frac{{3\left( {y - x} \right)\left( {y + x} \right)}}{{{x^2}{y^2}}} = 0\)

\( \Leftrightarrow \left( {x - y} \right)\left( {1 + \frac{{3x + 3y}}{{{x^2}{y^2}}}} \right) = 0\)

\( \Leftrightarrow \left[ \begin{array}{l}x - y = 0\\1 + \frac{{3x + 3y}}{{{x^2}{y^2}}} = 0\end{array} \right.\)

• TH1: x – y = 0 ⇔ x = y

Thay x = y vào phương trình \(2{\rm{x}} + y = \frac{3}{{{x^2}}}\) ta có:

\(2{\rm{x}} + x = \frac{3}{{{x^2}}} \Leftrightarrow 3{\rm{x}} = \frac{3}{{{x^2}}} \Leftrightarrow {x^3} = 1 \Leftrightarrow x = 1\)

Suy ra y = 1

• TH2: \(1 + \frac{{3x + 3y}}{{{x^2}{y^2}}} = 0\)

\( \Leftrightarrow \frac{{3x + 3y}}{{{x^2}{y^2}}} = - 1 \Leftrightarrow 3{\rm{x}} + 3y = - {x^2}{y^2} < 0\)  (1)

Ta có:

\(\left( {2{\rm{x}} + y} \right) + \left( {2y + x} \right) = \frac{3}{{{x^2}}} + \frac{3}{{{y^2}}}\)

\( \Leftrightarrow 3x + 3y = \frac{{3\left( {{y^2} + {x^2}} \right)}}{{{x^2}{y^2}}} > 0\)                   (2)

Từ (1) và (2) suy ra x, y ∈ ∅

Vậy (x, y) = (1, 1).