当x-y=m,y-z=n,求x²+y²+z²-xy-yz-zx的值? 解析出来！

Question

如题所述

Answer 1

x-y=m x*x-2xy+y*y=m*m
y-z=n y*y-2yz+z*z=n*n
x-z=m+n x*x-2xz+z*z=(m+n)^2
右边的式子相加的2（x²+y²+z²-xy-yz-zx）=m*m+n*n+（m+n）^2
得x²+y²+z²-xy-yz-zx=m²+n²+mn

Answer 2

x-z=(x-y)+(y-z)=m+n
x²+y²+z²-xy-yz-zx
=1/2[(x-y)²+(y-z)²+(x-z)²]
=1/2[m²+n²+(m+n)²]
=m²+n²+mn

Answer 3

由于X-Y=M,Y-Z=N，该两式相加可得X-Z=M+N,带入上式可得：
X^2+Y^2+Z^2-XY-YZ-ZX
=[(x-y)^2+(y-z)^2+(x-z)^2]/2 公式
=1/2*[M^2+N^2+(M+N)^2]
=1/2*(2M^2+2N^2+2MN)
=M^2+N^2+MN