Tìm ba số x,y,z liên tiếp của cấp số cộng biết yz=12 x^2=y^2+z^2

\[\begin{array}{l} x,\,\,y,\,\,\,z\,\,la\,\,\,3\,\,so\,\,lien\,\,tiep\,\,cua\,\,CSC\\ \Rightarrow x + z = 2y\,\,\,\,\,\,\left( 1 \right)\\ Lai\,\,co:\,\,\left\{ \begin{array}{l} yz = 12\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\\ {x^2} = {y^2} + {z^2}\,\,\,\,\,\,\left( 3 \right) \end{array} \right.\\ \Rightarrow \left( 2 \right) \Leftrightarrow y = \frac{{12}}{z}\\ \Rightarrow \left( 1 \right) \Leftrightarrow x + z = \frac{{24}}{z} \Leftrightarrow x = \frac{{24}}{z} - z\\ \Rightarrow \left( 3 \right) \Leftrightarrow {\left( {\frac{{24}}{z} - z} \right)^2} = \frac{{144}}{{{z^2}}} + {z^2}\\ \Leftrightarrow \frac{{576}}{{{z^2}}} - 48 + {z^2} = \frac{{144}}{{{z^2}}} + {z^2}\\ \Leftrightarrow \frac{{432}}{{{z^2}}} = 48\\ \Leftrightarrow {z^2} = 9\\ \Leftrightarrow \left[ \begin{array}{l} z = 3\\ z = - 3 \end{array} \right. \Rightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} x = 5\\ y = 4\\ z = 3 \end{array} \right.\\ \left\{ \begin{array}{l} x = - 5\\ y = - 4\\ z = - 3 \end{array} \right.. \end{array} \right. \end{array}\]