Question

cho x/2 =y/5=z/7.tính giá trị biểu thức A= x-y+z/x+2y-z

Answer 1

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

Vậy \(A=\frac{4}{5}\)

Answer 2

đặt x/2=y/6=z/7=k

suy ra    x-y+z/x+2-z  =  2k-5k+7k/2k10+7k = k(2-5+70/k(2+10-70 = 4/5

vậy A=4/5