Question

cho \(\dfrac{x}{2}\)=\(\dfrac{y}{5}=\dfrac{z}{7}\). tnh gia tri bieu thuc \(A=\dfrac{x-y+z}{x+2y-z}\)

Answer 1

Đặt \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=k\left(k\ne0\right)\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=5k\\z=7k\end{matrix}\right.\)

\(\Rightarrow A=\dfrac{2k-5k+7k}{2k+10k-7k}=\dfrac{k\left(2-5+7\right)}{k\left(2+10-7\right)}=\dfrac{4}{5}\)

Vậy A = \(\dfrac{4}{5}\)

Answer 2

Đặt \(\dfrac{x}{2}\) =\(\dfrac{y}{5}\) =\(\dfrac{z}{7}\) =k

=> A=\(\dfrac{x-y+x}{x+2y-z}\)=\(\dfrac{2k+5k-7k}{2k+10k-7k}\)=\(\dfrac{k\left(2-5+7\right)}{k\left(2+10-7\right)}\)=\(\dfrac{4}{5}\)

Vậy A= \(\dfrac{4}{5}\)

Answer 3

Đặt \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=k\)

\(\Rightarrow\)\(x=2k;y=5k;z=7k\)(1)

Thay (1) vào A ,ta có:

A=\(\dfrac{x-y+z}{x+2y-z}=\dfrac{2k-5k+7k}{2k+10k-7k}=\dfrac{4k}{5k}=\dfrac{4}{5}\)

Vậy A=\(\dfrac{4}{5}\)