Đáp án:
\[x = \dfrac{1}{2};\,\,\,y = \dfrac{5}{6};\,\,\,\,z = - \dfrac{5}{6}\]
Giải thích các bước giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\begin{array}{l}
\dfrac{{y + z + 1}}{x} = \dfrac{{x + z + 2}}{y} = \dfrac{{x + y - 3}}{z} = \dfrac{{\left( {y + z + 1} \right) + \left( {x + z + 2} \right) + \left( {x + y - 3} \right)}}{{x + y + z}}\\
= \dfrac{{2x + 2y + 2z}}{{x + y + z}} = \dfrac{{2.\left( {x + y + z} \right)}}{{x + y + z}} = 2\\
\Rightarrow \dfrac{{y + z + 1}}{x} = \dfrac{{x + z + 2}}{y} = \dfrac{{x + y - 3}}{z} = \dfrac{1}{{x + y + z}} = 2\\
\dfrac{1}{{x + y + z}} = 2 \Leftrightarrow x + y + z = \dfrac{1}{2} \Rightarrow \left\{ \begin{array}{l}
y + z = \dfrac{1}{2} - x\\
x + z = \dfrac{1}{2} - y\\
x + y = \dfrac{1}{2} - z
\end{array} \right.\\
\dfrac{{y + z + 1}}{x} = \dfrac{{x + z + 2}}{y} = \dfrac{{x + y - 3}}{z} = 2\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{{y + z + 1}}{x} = 2\\
\dfrac{{x + z + 2}}{y} = 2\\
\dfrac{{x + y - 3}}{z} = 2
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
y + z + 1 = 2x\\
x + z + 2 = 2y\\
x + y - 3 = 2z
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{1}{2} - x + 1 = 2x\\
\dfrac{1}{2} - y + 2 = 2y\\
\dfrac{1}{2} - z - 3 = 2z
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{3}{2} = 3x\\
\dfrac{5}{2} = 3y\\
- \dfrac{5}{2} = 3z
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{1}{2}\\
y = \dfrac{5}{6}\\
z = - \dfrac{5}{6}
\end{array} \right.
\end{array}\)
Vậy \(x = \dfrac{1}{2};\,\,\,y = \dfrac{5}{6};\,\,\,\,z = - \dfrac{5}{6}\)
