Question

Cho \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{11}\) . Tính giá trị của \(A=\dfrac{y+z-x}{x+z-y}\) 

Answer 1

Đặt \(\left\{{}\begin{matrix}\dfrac{x}{3}=k\\\dfrac{y}{4}=k\\\dfrac{z}{11}=k\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3k\\y=4k\\z=11k\end{matrix}\right.\)

Ta có: \(A=\dfrac{y+z-x}{x+z-y}\)

\(=\dfrac{4k+11k-3k}{3k+11k-4k}\)

\(=\dfrac{12k}{10k}=\dfrac{6}{5}\)

Answer 2

Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{11}=\dfrac{y+z-x}{12}=\dfrac{x+z-y}{10}\\ \Rightarrow\dfrac{y+z-x}{x+z-y}=\dfrac{12}{10}=\dfrac{6}{5}\)