已知方程组4x-3y-3z=0 x-3y-z=0求x的平方+xz分之xy+yz的值

4x-3y-3z=0 (1)
x-3y-z=0 (2)
(1)-(2)
3x-2z=0
z=(3/2)x
(1)-(2)×3
4x-3y-3x+9y=0
x-6y=0
y=(1/6)x
所以xz=x*(3/2)x=(3/2)x²
xy=x*(1/6)x=(1/6)x²
yz=(3/2)x*(1/6)x=(1/4)x²
所以原式=[(1/6)x²+(1/4)x²]/[x²+(3/2)x²]
=(5/12)x²/[(5/2)x²]
=(5/12)/(5/2)
=1/6