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Solución de sistemas de ecuaciones/ xyz=1; x-y-z=-1

xyz=1; x-y-z=-1

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Solución

Ha introducido [src] 
x*y*z = 1
zxy=1z x y = 1 
x - y - z = -1
−z+(x−y)=−1- z + \left(x - y\right) = -1 
-z + x - y = -1
Respuesta rápida
x1=z2−12−z(z+1)(z2−3z+4)2zx_{1} = \frac{z}{2} - \frac{1}{2} - \frac{\sqrt{z \left(z + 1\right) \left(z^{2} - 3 z + 4\right)}}{2 z}
=
z(z−1)−z(z+1)(z2−3z+4)2z\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

y1=−z2+12−z(z+1)(z2−3z+4)2zy_{1} = - \frac{z}{2} + \frac{1}{2} - \frac{\sqrt{z \left(z + 1\right) \left(z^{2} - 3 z + 4\right)}}{2 z}
=
z(1−z)−z(z+1)(z2−3z+4)2z\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
x2=z2−12+z(z+1)(z2−3z+4)2zx_{2} = \frac{z}{2} - \frac{1}{2} + \frac{\sqrt{z \left(z + 1\right) \left(z^{2} - 3 z + 4\right)}}{2 z}
=
z(z−1)+z(z+1)(z2−3z+4)2z\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

y2=−z2+12+z(z+1)(z2−3z+4)2zy_{2} = - \frac{z}{2} + \frac{1}{2} + \frac{\sqrt{z \left(z + 1\right) \left(z^{2} - 3 z + 4\right)}}{2 z}
=
z(1−z)+z(z+1)(z2−3z+4)2z\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
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