Đặt \(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}=k\left(k\ne0\right)\)
=>\(\left\{{}\begin{matrix}x=-4k\\y=-7k\\z=3k\end{matrix}\right.\)
Ta có P =\(\dfrac{-2\cdot\left(-4k\right)+\left(-7k\right)+5\cdot3k}{2\cdot\left(-4k\right)-3\left(-7k\right)-6\left(3k\right)}\)=\(\dfrac{8k+\left(-7k\right)+15k}{-8k+21k-18k}\)=
\(\dfrac{k\cdot\left(8+\left(-7\right)+15\right)}{k.\left(-8+21-18\right)}=\dfrac{-16}{5}\)
Vậy P= \(\dfrac{-16}{5}\)
Theo đề ta có:
\(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}\)
Đặt k cho biểu thức trên
=>\(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}\) =k
=> \(\left[{}\begin{matrix}\dfrac{x}{-4}=k\\\dfrac{y}{-7}=k\\\dfrac{z}{3}=k\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\left(-4\right).k\\y=\left(-7\right).k\\z=3.k\end{matrix}\right.\)
Thay \(\left[{}\begin{matrix}x=\left(-4\right).k\\y=\left(-7\right).k\\z=3.k\end{matrix}\right.\) vào biểu thức \(P=\dfrac{-2x+y+5z}{2x-3y-6z}\)
Ta được:
\(P=\dfrac{-2.\left(-4.k\right)+\left(-7.k\right)+5\left(3.k\right)}{2\left(-4.k\right)-3\left(-7.k\right)-6\left(3.k\right)}\)
=> \(P=\dfrac{8.k+\left(-7.k\right)+15.k}{-8.k+21.k-18.k}\)
=> \(P=\dfrac{k.\left(8+-7+15\right)}{k.\left(-8+21-18\right)}\)
=> P= \(-\dfrac{16}{5}\)
Vậy:....................
