Cho x; y; z khác ± 1 và xy + yz + xz = 1. Chọn câu đúng?

* Hướng dẫn giải

Ta có

$\frac{\mathrm{x}}{1-{\mathrm{x}}^2}+\frac{\mathrm{y}}{1-{\mathrm{y}}^2}+\frac{\mathrm{z}}{1-{\mathrm{z}}^2}$

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

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

=$\frac{\mathrm{x}-{\mathrm{xz}}^2-{\mathrm{xy}}^2+{\mathrm{xy}}^2{\mathrm{z}}^2+\mathrm{y}-{\mathrm{x}}^2\mathrm{y}-{\mathrm{yz}}^2+{\mathrm{yz}}^2{\mathrm{x}}^2+\mathrm{z}-{\mathrm{zx}}^2-{\mathrm{zy}}^2+{\mathrm{zx}}^2{\mathrm{y}}^2}{(1-{\mathrm{x}}^2\left)\right(1-{\mathrm{y}}^2\left)\right(1-{\mathrm{z}}^2)}$

$=\frac{(\mathrm{x}-\mathrm{y}{\mathrm{x}}^2-{\mathrm{x}}^2\mathrm{z})+(\mathrm{y}-{\mathrm{xy}}^2-{\mathrm{zy}}^2)+(\mathrm{z}-{\mathrm{xz}}^2-{\mathrm{yz}}^2)+(\mathrm{x}{\mathrm{y}}^2{\mathrm{z}}^2+{\mathrm{yz}}^2{\mathrm{x}}^2+{\mathrm{zx}}^2{\mathrm{y}}^2)}{(1-{\mathrm{x}}^2\left)\right(1-{\mathrm{y}}^2\left)\right(1-{\mathrm{z}}^2)}$

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

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

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

Đáp án cần chọn là: C