$\frac{3x^{-2}+2y^{-2}}{2x^{-2}+3y^{-2}}=\frac{\frac{3}{x^2}+\frac{2}{y^2}}{\frac{2}{x^2}+\frac{3}{y^2}}=\frac{\frac{3y^2+2x^2}{x^2{y}^2}}{\frac{2y^2+3x^2}{x^2{y}^2}}=\frac{3y^2+2x^2}{x^2{y}^2}\ast \frac{x^2{y}^2}{2y^2+3x^2}=\frac{3y^2+2x^2}{2y^2+3x^2}$
при $\frac{x}{y}=2^{-1}=\frac{1}{2}$:
y = 2x, тогда:
$\frac{3y^2+2x^2}{2y^2+3x^2}=\frac{3\ast (2x{)}^2+2x^2}{2\ast (2x{)}^2+3x^2}=\frac{12x^2+2x^2}{8x^2+3x^2}=\frac{14x^2}{11x^2}=\frac{14}{11}=1\frac{3}{11}$

$\frac{3x^{-2}-2y^{-2}}{2x^{-2}-3y^{-2}}=\frac{\frac{3}{x^2}-\frac{2}{y^2}}{\frac{2}{x^2}-\frac{3}{y^2}}=\frac{\frac{3y^2-2x^2}{x^2{y}^2}}{\frac{2y^2-3x^2}{x^2{y}^2}}=\frac{3y^2-2x^2}{x^2{y}^2}\ast \frac{x^2{y}^2}{2y^2-3x^2}=\frac{3y^2-2x^2}{2y^2-3x^2}$
при $\frac{x}{y}=3^{-1}=\frac{1}{3}$:
y = 3x, тогда:
$\frac{3y^2-2x^2}{2y^2-3x^2}=\frac{3\ast (3x{)}^2-2x^2}{2\ast (3x{)}^2-3x^2}=\frac{25x^2}{15x^2}=\frac{25}{15}=\frac{5}{3}=1\frac{2}{3}$