Question

cho \(\frac{x}{2}\) = \(\frac{y}{5}\) =\(\frac{z}{7}\) tính giá trị của biểu thức A=\(\frac{x-y+z}{x+2y-z}\) giúp mink nha

Answer 1

Ta có: \(\frac{x}{2}=\frac{y}{5}=\frac{z}{7}\)

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

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

Theo đề ta có:

\(A=\frac{x-y+z}{x+2y-z}=\frac{2k-5k+7k}{2k+2\left(5k\right)-7k}\)

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

\(A=\frac{4k}{5k}=\frac{4}{5}\)

Vậy \(A=\frac{4}{5}\)

Answer 2

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

\(\Rightarrow\hept{\begin{cases}x=2k\\y=5k\\z=7k\end{cases}}\)

Thay vào \(\frac{x-y+z}{x+2y-z}\)ta có :

\(A=\frac{2k-5k+7k}{2k+2.5k-7k}=\frac{\left(2-5+7\right)k}{2k+10k-7k}=\frac{4k}{\left(2+10-7\right)k}=\frac{4k}{5k}=\frac{4}{5}\)

Vậy \(A=\frac{4}{5}\)