Đáp án + Giải thích các bước giải:

Bài $\textbf{9}.$

$\dfrac{x}{2 - x} + \dfrac{y}{2 - y} = 1$

$\dfrac{x - 2 + 2}{2 - x} + \dfrac{y - 2 + 2}{2 - y} = 1$

$-1 + \dfrac{2}{2 - x} - 1 + \dfrac{2}{2 - y} = 1$

$\dfrac{2}{2 - x} + \dfrac{2}{2 - y} = 3$

$\dfrac{4 - 2y + 4 - 2x}{(2 - x)(2 - y)} = 3$

$\dfrac{8 - 2x - 2y}{(2 - x)(2 - y)} = 3$

$8 - 2x - 2y = 3(2 - x)(2 - y)$

$8 - 2x - 2y = 12 - 6x - 6y + 3xy$

$3xy - 4x - 4y + 4 = 0$

$4x + 4y = 3xy + 4$

$(4x + 4y)^2 = (3xy + 4)^2$

$16x^2 + 32xy + 16y^2 = 9x^2y^2 + 24xy + 16$

$16x^2 + 8xy + 16y^2 - 16= 9x^2y^2$

$\begin{array}{l} \textbf{T} &= \sqrt{16x^2 + 8xy + 16y^2 - 16} - 3xy
\\ &= \sqrt{9x^2y^2} - 3xy
\\ &= 3xy - 3xy (0 < x, y < 2)
\\ &= 0\end{array}$