Sr Examen
xy=1; z=x+y
Solución
Respuesta rápida
$$x_{1} = \frac{z}{2} - \frac{\sqrt{\left(z - 2\right) \left(z + 2\right)}}{2}$$
=
$$\frac{z}{2} - \frac{\sqrt{z^{2} - 4}}{2}$$
=
$$y_{1} = \frac{z}{2} + \frac{\sqrt{\left(z - 2\right) \left(z + 2\right)}}{2}$$
=
$$\frac{z}{2} + \frac{\sqrt{z^{2} - 4}}{2}$$
=
=
$$\frac{z}{2} - \frac{\sqrt{z^{2} - 4}}{2}$$
=
0.5*z - 0.5*((2 + z)*(-2 + z))^0.5
$$y_{1} = \frac{z}{2} + \frac{\sqrt{\left(z - 2\right) \left(z + 2\right)}}{2}$$
=
$$\frac{z}{2} + \frac{\sqrt{z^{2} - 4}}{2}$$
=
0.5*z + 0.5*((2 + z)*(-2 + z))^0.5
$$x_{2} = \frac{z}{2} + \frac{\sqrt{\left(z - 2\right) \left(z + 2\right)}}{2}$$
=
$$\frac{z}{2} + \frac{\sqrt{z^{2} - 4}}{2}$$
=
$$y_{2} = \frac{z}{2} - \frac{\sqrt{\left(z - 2\right) \left(z + 2\right)}}{2}$$
=
$$\frac{z}{2} - \frac{\sqrt{z^{2} - 4}}{2}$$
=
=
$$\frac{z}{2} + \frac{\sqrt{z^{2} - 4}}{2}$$
=
0.5*z + 0.5*((2 + z)*(-2 + z))^0.5
$$y_{2} = \frac{z}{2} - \frac{\sqrt{\left(z - 2\right) \left(z + 2\right)}}{2}$$
=
$$\frac{z}{2} - \frac{\sqrt{z^{2} - 4}}{2}$$
=
0.5*z - 0.5*((2 + z)*(-2 + z))^0.5