Question

已知xyz均不为零,且x+2y-z=0,7x-y-z=0,求x+y+z除以2x-y-z的值

Answer 1

因为x+2y-z=0，7x-y-z=0

两式相减，得：6x-3y=0，所以y=2x
代入x+2y-z=0中，得：x+4x-z=0，那么z=5x
那么(x+y+z)÷(2x-y-z)=(x+2x+5x)÷(2x-2x-5x)
=(8x)÷(-5x)
=-8/5

Answer 2

x+2y-z=0①
7x-y-z=0②

②-①，得
6x-3y=0
y=2x
代入①，得
x+4x-z=0
z=5x
所以
x+y+z除以2x-y-z

=(x+2x+5x)/(2x-x-5x)
=-8/4
=-2

Answer 3

x+2y-z=0,7x-y-z=0,这两个式子相减可得：6x-3y=0，即y=2x，代入x+2y-z=0可得：z=5x，所以
(x+y+z)/(2x-y-z)=(x+2x+5x)/(2x-2x-5x)=-8/5

Answer 4

7x-y-z=0,可知y+z=7x,-y-z=-7x;
x+y+z=8x,2x-y-z=-5x
x+y+z除以2x-y-z=-8/5

Answer 5

由x+2y-z=0,7x-y-z=0可得，
z=5x,y=7x
∴原式=13x除以-10x
=1.3

Answer 6

-8/5