Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{\left(2x-2\right)+\left(2y-6\right)-\left(z-3\right)}{4+9-4}=\dfrac{\left(2x+2y-z\right)-2-6+3}{9}=\dfrac{45}{9}=5\)
\(\Rightarrow\left\{{}\begin{matrix}x=5.2+1\\y=5.3+2\\z=5.4+3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=11\\y=17\\z=23\end{matrix}\right.\)
Vậy, x = 11; y = 17; z = 23
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2.\left(x-1\right)+3.\left(y-2\right)-\left(z-3\right)}{\left(2.2\right)+\left(3.3\right)-4}=\dfrac{2x-2+3y-6-z+3}{4+9-4}\)
\(=\dfrac{\left(2x+3y-z\right)-5}{9}=\dfrac{50-5}{9}=\dfrac{45}{9}=5\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=5\\\dfrac{y-2}{3}=5\\\dfrac{z-3}{4}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=5.2\\y-2=5.3\\z-3=5.4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=10\\y-2=15\\z-3=20\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=10+1\\y=15+2\\z=20+3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=11\\y=17\\z=23\end{matrix}\right.\)
Vậy .......................
