Cho \(x;\,y;\,z\, \ne  \pm 1\) và \(xy + yz + x{\rm{z}} = 1\). Chọn câu đúng?

\(\begin{array}{l}\frac{x}{{1 - {x^2}}} + \frac{y}{{1 - {y^2}}} + \frac{z}{{1 - {z^2}}}\\ = \frac{{x\left( {1 - {y^2}} \right)\left( {1 - {z^2}} \right) + y\left( {1 - {x^2}} \right)\left( {1 - {z^2}} \right) + z\left( {1 - {x^2}} \right)\left( {1 - {y^2}} \right)}}{{\left( {1 - {x^2}} \right)\left( {1 - {y^2}} \right)\left( {1 - {z^2}} \right)}}\\ = \frac{{x\left( {1 - {y^2} - {z^2} + {y^2}{z^2}} \right) + y\left( {1 - {x^2} - {z^2} + {x^2}{z^2}} \right) + z\left( {1 - {x^2} - {y^2} + {x^2}{y^2}} \right)}}{{\left( {1 - {x^2}} \right)\left( {1 - {y^2}} \right)\left( {1 - {z^2}} \right)}}\\ = \frac{{x - x{y^2} - x{z^2} + x{y^2}{z^2} + y - {x^2}y - y{z^2} + {x^2}y{z^2} + z - {x^2}z - {y^2}z + {x^2}{y^2}z}}{{\left( {1 - {x^2}} \right)\left( {1 - {y^2}} \right)\left( {1 - {z^2}} \right)}}\\ = \frac{{\left( {x - {x^2}y - {x^2}z} \right) + \left( {y - x{y^2} - {y^2}z} \right) + \left( {z - x{{\rm{z}}^2} - y{z^2}} \right) + \left( {x{y^2}{z^2} + {x^2}y{z^2} + {x^2}{y^2}z} \right)}}{{\left( {1 - {x^2}} \right)\left( {1 - {y^2}} \right)\left( {1 - {z^2}} \right)}}\\ = \frac{{x\left( {1 - xy - x{\rm{z}}} \right) + y\left( {1 - xy - yz} \right) + z\left( {1 - x{\rm{z}} - yz} \right) + xyz\left( {yz + x{\rm{z}} + xy} \right)}}{{\left( {1 - {x^2}} \right)\left( {1 - {y^2}} \right)\left( {1 - {z^2}} \right)}}\\ = \frac{{x.yz + y.x{\rm{z}} + z.xy + xyz.1}}{{\left( {1 - {x^2}} \right)\left( {1 - {y^2}} \right)\left( {1 - {z^2}} \right)}}\\ = \frac{{4xyz}}{{\left( {1 - {x^2}} \right)\left( {1 - {y^2}} \right)\left( {1 - {z^2}} \right)}}\end{array}\)