Đáp án:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
$\begin{array}{l}
\dfrac{{y + z + 1}}{x} = \dfrac{{z + x + 2}}{y} = \dfrac{{x + y - 3}}{z} = \dfrac{1}{{x + y + z}}\\
= \dfrac{{y + z + 1 + z + x + 2 + x + y - 3}}{{x + y + z}}\\
= \dfrac{{2x + 2y + 2z}}{{x + y + z}} = 2\\
\Rightarrow \left\{ \begin{array}{l}
x + y + z = \dfrac{1}{2}\\
y + z + 1 = 2x\\
z + x + 2 = 2y\\
x + y - 3 = 2z
\end{array} \right.\left( * \right)\\
\Rightarrow \left\{ \begin{array}{l}
x + y = \dfrac{1}{2} - z\left( 1 \right)\\
y + z = \dfrac{1}{2} - x\left( 2 \right)\\
z + x = \dfrac{1}{2} - y\left( 3 \right)
\end{array} \right.\\
Thay\,\left( 1 \right);\left( 2 \right);\left( 3 \right)\,vào\,\left( * \right)\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{1}{2} - z - 3 = 2z\\
\dfrac{1}{2} - x + 1 = 2x\\
\dfrac{1}{2} - y + 2 = 2y
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
3z = - \dfrac{5}{2}\\
3x = \dfrac{3}{2}\\
3y = \dfrac{5}{2}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
z = - \dfrac{5}{6}\\
x = \dfrac{1}{2}\\
y = \dfrac{5}{6}
\end{array} \right.\\
Vậy\,x = \dfrac{1}{2};y = \dfrac{5}{6};z = - \dfrac{5}{6}
\end{array}$
