Solución problema 1

Enunciado problema 1

Como

$$\{\begin{array}{cc}\begin{array}{c}x+y=\sqrt{5}\\ x-y=\sqrt{3}\end{array}& x=\frac{\sqrt{5}+\sqrt{3}}{2};y=\frac{\sqrt{5}-\sqrt{3}}{2}\end{array}$$

Por tanto

$$\frac{x^2}{y^2}-\frac{y^2}{x^2}=\frac{x^4-y^4}{{\left(xy\right)}^2}=\frac{\left(x^2+y^2\right)\left(x^2-y^2\right)}{{\left(xy\right)}^2}$$

Como

$$xy=\frac{5-3}{4}=\frac{1}{2}$$

$$x^2-y^2=\left(x+y\right)\left(x-y\right)=\sqrt{15}$$

$$x^2+y^2={\left(x+y\right)}^2-2xy=5-2\cdot \frac{1}{2}=4$$

Entonces en

$$\frac{4\sqrt{15}}{{\left(\frac{1}{2}\right)}^2}=16\sqrt{15}$$