Question

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

Answer 1

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