Ta có: \(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)
\(\Leftrightarrow\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{x-1}{2}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}=\dfrac{\left(x-1\right)-\left(2y-4\right)+\left(3z-9\right)}{2-6+12}\)
\(=\dfrac{x-1-2y+4+3z-9}{2-6+12}\)\(=\dfrac{\left(x-2y+3z\right)-\left(1-4+9\right)}{8}\)
\(=\dfrac{-10-6}{8}\)\(=\dfrac{-16}{8}=-2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=-2\Rightarrow x-1=-4\Rightarrow x=-3\\\dfrac{2y-4}{6}=-2\Rightarrow2y-4=-12\Rightarrow2y=-8\Rightarrow y=-4\\\dfrac{3z-9}{12}=-2\Rightarrow3z-9=-24\Rightarrow3z=-15\Rightarrow z=-5\end{matrix}\right.\)
Vậy 3 số \(x,y,z\) cần tìm lần lượt là \(\left(-3\right),\left(-4\right),\left(-5\right)\)
