问题描述:
已知x^3+y^3+z^3=96,xyz=4,x^2+y^2+z^2-xy+xz+yz=12,则x+y-z的值是多少?最佳答案:
答案是4.
3√(96+2*12+(2+(-2)+(-2))*4-4*12)
=3√(96+24-8-48)
=3√64
=4
算了一小时^_^
已知x^3+y^3+z^3=96,xyz=4,x^2+y^2+z^2-xy+xz+yz=12,则x+y-z的值是多少?
问题描述:
已知x^3+y^3+z^3=96,xyz=4,x^2+y^2+z^2-xy+xz+yz=12,则x+y-z的值是多少?答案是4.
3√(96+2*12+(2+(-2)+(-2))*4-4*12)
=3√(96+24-8-48)
=3√64
=4
算了一小时^_^