Bài làm:

Đặt \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=K\hept{\begin{cases}x=3K\\y=4K\\z=5K\end{cases}}\)

\(\Rightarrow\frac{x+y-z}{x+2y-z}\)

\(\Rightarrow\frac{3K+4K-5K}{3K+2\left(4K\right)-5K}\)

\(\Rightarrow\frac{3K+4K-5K}{3K+8K-5K}\)

\(\Rightarrow\frac{K\left(3+4-5\right)}{K\left(3+8-5\right)}\)

\(\Rightarrow\frac{3+4-5}{3+8-5}=\frac{2}{6}\)

Vậy biểu thức B = \(\frac{2}{6}\)

khó quá ?????????????????????????????????????????????????????

X/3=Y/4=Z/5 -> X=3 , Y=4 , Z=5 -> 3+4-5/3+(2.4)-5=1/4

1/4 nha bn

a) A = x - y + z + z + y + x - 2y

A = (x + x) + (-y + y) + (z + z) - 2y

A = 2x + 0 + 2z - 2y 

A = 2 .(x + z - y)

b) Thay x = 3 ; y = -1 ; z = 2 vào biểu thức A , ta được :

A = 2 .[3 + 2 - (-1)]

A = 12

Vậy A = 12

Chúc bạn học tốt !

Ta có:

\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=\frac{x+y+z}{3+4+5}=\frac{x+y+z}{x+2y-z}.\left(\text{dãy tỉ số bằng nhau}\right)\)

\(\Rightarrow3+4+5=12=x+2y-z\)

\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=\frac{x+2y-z}{3+8-5}=\frac{12}{6}=2\)

=> x/3 =2 =>x=6

=> x/4=2 => x=8

=> x/5 =2 => x=10

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 \(A=\frac{4}{5}\)

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

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

=>x=2k; y=5k; z=7k

Theo bài ra ta có:

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

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

theo bài ra ta có:

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

áp dụng tính chất dãy tỉ số bằng nhau ta có:

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

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

theo tính chất tỉ lệ thức ta có;

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

vậy A = \(\frac{4}{5}\)

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

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

Ta có: \(A=\frac{x-y+z}{x+2y-z}=\frac{2k-5k+7k}{2k+2\left(5k\right)-7k}=\frac{k\left(2-5+7\right)}{2k+10k-7k}=\frac{k4}{\left(2+10-7\right)k}=\frac{4}{5}\)

Vậy \(A=\frac{4}{5}\)