Câu `6`

`x/2=y/3 => x/8=y/12`

`5y=4z => y/4=z/5 => y/12=z/15`

`=>x/8=y/12=z/15`

`\text{Áp dụng t/c dãy tỉ số bằng nhau ta có :}`

`x/8=y/12=z/15=(2x-y+z)/(2*8-12+15)=(x+y-z)/(8+12-15)`

`to (2x-y+z)/(x+y-z)=(2*8-12+15)/(8+12-15)=(16-12+15)/(20-15)=(4+15)/5=19/5`

Vậy `(2x-y+z)/(x+y-z)=19/5`