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

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

Ta có: \(A=\frac{x-y+z}{x+2y-z}\)

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

Vậy \(A=\frac{4}{5}\)

+ \(\frac{x}{2}=\frac{y}{5}=\frac{z}{7}=\frac{x-y+z}{2-5+7}=\frac{x-y+x}{4}\Rightarrow x-y+x=2x\)

+ \(\frac{x}{2}=\frac{2y}{10}=\frac{z}{7}=\frac{x+2y-z}{2+10-7}=\frac{x+2y-z}{5}\Rightarrow x+2y-z=\frac{5x}{2}\)

\(\Rightarrow\frac{x-y+z}{x+2y-z}=\frac{2x.2}{5x}=\frac{4}{5}\)

4/5 bạn nhé

+ x2 =y5 =z7 =x−y+z2−5+7 =x−y+x4 ⇒x−y+x=2x

+ x2 =2y10 =z7 =x+2y−z2+10−7 =x+2y−z5 ⇒x+2y−z=5x2 

⇒x−y+zx+2y−z =2x.25x =45 

Ta có \(\frac{x}{5}=\frac{y}{5}=\frac{z}{7}=\frac{2y}{10}\)

Áp dụng tính chất dãy tỉ số bằng nhau tao có

\(\frac{x}{5}=\frac{y}{5}=\frac{z}{7}=\frac{x-y+z}{5-5+7}=\frac{x-y+z}{7}\left(1\right)\)

\(\frac{x}{5}=\frac{2y}{10}=\frac{z}{7}=\frac{x+2y-z}{5+10-7}=\frac{x+2y-z}{8}\left(2\right)\)

Từ (1) và (2) ta được \(\frac{x-y+z}{7}=\frac{x+2y-z}{8}\Rightarrow\frac{x-y+z}{x+2y-z}=\frac{7}{8}\)

Vậy A= \(\frac{7}{8}\)

Study Well !

đợi mk đi có việc đã , xong sẽ quay lại giải giùm bn nghe Lê Trần Hoàng Oanh

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

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

Ta có: \(A=\frac{x-y+z}{x+2y-z}=\frac{2k-5k+7k}{2k+10k-7k}=\frac{k.\left(2-5+7\right)}{k\left(2+10-7\right)}=\frac{4k}{5k}=\frac{4}{5}\)

Vậy \(A=\frac{4}{5}\)

\(Đặt x / 2 = y / 5 = z / 7 = k \)

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

\(A = ( 4x - y + z ) / ( x + 2y - z )\)

\(A = ( 4 . 2k - 5k + 7k ) / ( 2k + 2 . 5k - 7k ) \)

\(A = ( 8k - 5k + 7k ) / ( 2k + 10k - 7k )\)

\(A = 10 k / 5k\)

\(A = 2\)

đặt x=2k ,y=5k, z=7k

=>A=2k-5k+7k/2k+10k-7k

      =(2-5+7)k/(2+10-7)k

     =4k/5k =4/5