(初一)数学题分享

解:xy+yz+xz=(x+y+z)²-(x²+y²+z²)/2=－6

x²y²+y²z²+x²z²=(xy+yz+xz)²－2xyz(x+y+z)=32

则原式=1/－6－yz－xz+2z+1/－6－xy－xz+2z+1/6－xy－yz+2y

=1/－6－z(x+y－2)+1/－6－x(y+z－2)+1/－6－y(x+z－2)

=1/z²－6+1/x²－6+1/y²－6

=(x²-6)(y²－6)+(z²－6)(y²－6)+(x²－6)(z²－6)/(x²－6)(y²－6)(z²－6)

=x²y²+y²z²+x²z²－12(x²+y²+z²)+108/x²y²z²－6(x²y²+y²z²+x²z²)+36(x²+y²+z²)－216

=32－12×16+108/1－6×32+36×16－216

=－4/13