Sr Examen
xyz=1; x-y-z=-1
Solución
Ha introducido
[src]
x*y*z = 1
$$z x y = 1$$
x - y - z = -1
$$- z + \left(x - y\right) = -1$$
-z + x - y = -1
Respuesta rápida
$$x_{1} = \frac{z}{2} - \frac{1}{2} - \frac{\sqrt{z \left(z + 1\right) \left(z^{2} - 3 z + 4\right)}}{2 z}$$
=
$$\frac{z \left(z - 1\right) - \sqrt{z \left(z + 1\right) \left(z^{2} - 3 z + 4\right)}}{2 z}$$
=
$$y_{1} = - \frac{z}{2} + \frac{1}{2} - \frac{\sqrt{z \left(z + 1\right) \left(z^{2} - 3 z + 4\right)}}{2 z}$$
=
$$\frac{z \left(1 - z\right) - \sqrt{z \left(z + 1\right) \left(z^{2} - 3 z + 4\right)}}{2 z}$$
=
=
$$\frac{z \left(z - 1\right) - \sqrt{z \left(z + 1\right) \left(z^{2} - 3 z + 4\right)}}{2 z}$$
=
-0.5 + 0.5*z - 0.5*(z*(1 + z)*(4 + z^2 - 3*z))^0.5/z
$$y_{1} = - \frac{z}{2} + \frac{1}{2} - \frac{\sqrt{z \left(z + 1\right) \left(z^{2} - 3 z + 4\right)}}{2 z}$$
=
$$\frac{z \left(1 - z\right) - \sqrt{z \left(z + 1\right) \left(z^{2} - 3 z + 4\right)}}{2 z}$$
=
0.5 - 0.5*z - 0.5*(z*(1 + z)*(4 + z^2 - 3*z))^0.5/z
$$x_{2} = \frac{z}{2} - \frac{1}{2} + \frac{\sqrt{z \left(z + 1\right) \left(z^{2} - 3 z + 4\right)}}{2 z}$$
=
$$\frac{z \left(z - 1\right) + \sqrt{z \left(z + 1\right) \left(z^{2} - 3 z + 4\right)}}{2 z}$$
=
$$y_{2} = - \frac{z}{2} + \frac{1}{2} + \frac{\sqrt{z \left(z + 1\right) \left(z^{2} - 3 z + 4\right)}}{2 z}$$
=
$$\frac{z \left(1 - z\right) + \sqrt{z \left(z + 1\right) \left(z^{2} - 3 z + 4\right)}}{2 z}$$
=
=
$$\frac{z \left(z - 1\right) + \sqrt{z \left(z + 1\right) \left(z^{2} - 3 z + 4\right)}}{2 z}$$
=
-0.5 + 0.5*z + 0.5*(z*(1 + z)*(4 + z^2 - 3*z))^0.5/z
$$y_{2} = - \frac{z}{2} + \frac{1}{2} + \frac{\sqrt{z \left(z + 1\right) \left(z^{2} - 3 z + 4\right)}}{2 z}$$
=
$$\frac{z \left(1 - z\right) + \sqrt{z \left(z + 1\right) \left(z^{2} - 3 z + 4\right)}}{2 z}$$
=
0.5 - 0.5*z + 0.5*(z*(1 + z)*(4 + z^2 - 3*z))^0.5/z