Question

Answer 1

Đặt \(\frac{x}{-4}=\frac{y}{-7}=\frac{z}{3}=k\)

=>x=-4k; y=-7k; z=3k

\(A=\frac{-2x+y+5z}{2x-3y-6z}\)

\(=\frac{-2\cdot\left(-4k\right)+\left(-7k\right)+5\cdot3k}{2\cdot\left(-4k\right)-3\cdot\left(-7k\right)-6\cdot3k}\)

\(=\frac{8k-7k+15k}{-8k+21k-18k}=\frac{8-7+15}{-8+21-18}=\frac{1+15}{13-18}=\frac{16}{-5}=-\frac{16}{5}\)

=>Chọn D