a)
+) Ta có: \(x=\dfrac{y}{2}=\dfrac{z}{3}=k\Rightarrow\left\{{}\begin{matrix}x=k\\y=2k\\z=3k\end{matrix}\right.\)
+) Đặt: \(4x-3y+2z=36\)
\(\Rightarrow4.k+3.2k-2.3k=36\)
\(\Rightarrow4k+6k-6k=36\)
\(\Rightarrow4k=36\\\)
\(\Rightarrow k=36:4\)
\(\Rightarrow k=9\)
+) Thay vào cách đặt: \(\left\{{}\begin{matrix}x=9\\y=2.9\\z=3.9\end{matrix}\right.\)
Vậy: \(\left\{{}\begin{matrix}x=9\\y=18\\z=27\end{matrix}\right.\)
b)
+) Ta có: \(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{7}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=4k\\z=7k\end{matrix}\right.\)
+) Đặt: \(2x-3z=44\)
\(\Rightarrow2.5k-3.7k=44\)
\(\Rightarrow-11k=44\)
\(\Rightarrow k=44:\left(-11\right)\)
\(\Rightarrow k=-4\)
+) Thay vào cách đặt :\(\left\{{}\begin{matrix}x=5.\left(-4\right)\\y=4.\left(-4\right)\\z=7.\left(-4\right)\end{matrix}\right.\)
+) Vậy : \(\left\{{}\begin{matrix}x=-20\\y=-16\\z=-28\end{matrix}\right.\)
d)
+) Ta có: \(\dfrac{y}{12}=\dfrac{-x}{5}=\dfrac{z}{11}=k\Rightarrow\left\{{}\begin{matrix}y=12k\\x=-5k\\z=11k\end{matrix}\right.\)
* \(\dfrac{-x}{5}\) đổi ngược lại thành \(\dfrac{x}{-5}\)
+) Đặt: \(5y-2z=114\)
\(\Rightarrow5.12k-2.11k=114\)
\(\Rightarrow38k=114\)
\(\Rightarrow k=114:38\)
\(\Rightarrow k=3\)
+) Thay vào cách đặt: \(\left\{{}\begin{matrix}y=12.3\\x=\left(-5\right).3\\z=11.3\end{matrix}\right.\)
Vậy: \(\left\{{}\begin{matrix}y=36\\x=-15\\z=33\end{matrix}\right.\)
Mình giải vậy thôi bạn nhá!!!