\(\dfrac{x-1}{2}\) hay \(x\) - \(\dfrac{1}{2}\) vậy em?
\(\dfrac{x-1}{2}\) = \(\dfrac{y-2}{3}\) = \(\dfrac{z-3}{4}\)
\(\dfrac{2.\left(x-1\right)}{2.2}\) = \(\dfrac{3.\left(y-2\right)}{3.3}\) = \(\dfrac{z-3}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2.\left(x-1\right)}{4}\) = \(\dfrac{2x-2+3y-6-z+3}{4+9-4}\) = \(\dfrac{2x+3y-z-5}{9}\)=\(\dfrac{50-5}{9}\)= 5
⇒ \(\left\{{}\begin{matrix}\dfrac{x-1}{2}=5\\\dfrac{y-2}{3}=5\\\dfrac{z-3}{4}=5\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}x=11\\y=17\\z=23\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)\) = (11;17;23)
