已知x⼀2=y⼀3=z⼀4≠0,求(x-2y)⼀z的值

2025-03-04 12:13:48
推荐回答(3个)
回答1:

回答2:

x/2=y/3=z/4≠0可推出y=3/2x,z=2x所以2x+3y-z=2x+9/2x-2x=9/2x       x-2y+3z=x-3x+6x=4x(2x+3y-z)/(x-2y+3z)=9/8

回答3:

设x/3=y/4=z/5=k 所以 x=3k y=4k z=5k 所以 (x+y-z)/(x-y+z) =(3k+4k-5k)/(3k-4k+5k) =2k/4k =1/2 祝学习进步