Cho \(x;y;z\) khác \( \pm 1\) và \(xy + yz + xz = 1.\) Chọn câu đúng.

Ta có

\(\dfrac{x}{{1 - {x^2}}} + \dfrac{y}{{1 - {y^2}}} + \dfrac{z}{{1 - {z^2}}}\)

\( = \dfrac{{x\left( {1 - {y^2}} \right)\left( {1 - {z^2}} \right) + y\left( {1 - {x^2}} \right)\left( {1 - {z^2}} \right) + z\left( {1 - {x^2}} \right)\left( {1 - {y^2}} \right)}}{{\left( {1 - {x^2}} \right)\left( {1 - {y^2}} \right)\left( {1 - {z^2}} \right)}}\)

\( = \dfrac{{x\left( {1 - {z^2} - {y^2} + {z^2}{y^2}} \right) + y\left( {1 - {x^2} - {z^2} + {x^2}{z^2}} \right) + z\left( {1 - {x^2} - {y^2} + {x^2}{y^2}} \right)}}{{\left( {1 - {x^2}} \right)\left( {1 - {y^2}} \right)\left( {1 - {z^2}} \right)}}\)

\( = \dfrac{{x - x{z^2} - x{y^2} + x{y^2}{z^2} + y - y{x^2} - y{z^2} + y{z^2}{x^2} + z - z{x^2} - z{y^2} + z{x^2}{y^2}}}{{\left( {1 - {x^2}} \right)\left( {1 - {y^2}} \right)\left( {1 - {z^2}} \right)}}\)

\( = \dfrac{{\left( {x - y{x^2} - x{z^2}} \right) + \left( {y - x{y^2} - z{y^2}} \right) + \left( {z - x{z^2} - y{z^2}} \right) + \left( {x{y^2}{z^2} + y{z^2}{x^2} + z{x^2}{y^2}} \right)}}{{\left( {1 - {x^2}} \right)\left( {1 - {y^2}} \right)\left( {1 - {z^2}} \right)}}\)

\( = \dfrac{{x\left( {1 - xy - xz} \right) + y\left( {1 - xy - yz} \right) + z\left( {1 - xz - zy} \right) + xyz\left( {yz + xz + xy} \right)}}{{\left( {1 - {x^2}} \right)\left( {1 - {y^2}} \right)\left( {1 - {z^2}} \right)}}\)

\( = \dfrac{{x.yz + y.xz + z.xy + xyz}}{{\left( {1 - {x^2}} \right)\left( {1 - {y^2}} \right)\left( {1 - {x^2}} \right)}}\)

\( = \dfrac{{4xyz}}{{\left( {1 - {x^2}} \right)\left( {1 - {y^2}} \right)\left( {1 - {z^2}} \right)}}\)