Question

若x-y=m,y-z=n,试求x²+y²+z²-xy-yz-xz的值

若x-y=m,y-z=n,试求x²+y²+z²-xy-yz-xz的值

Answer 1

x-y = m => x²+y²-2xy=m²
y-z = n => y²+z²-2yz=n²
x-z =m-n => x²+z²-2xz=(m+n)²

x²+y²-2xy+y²+z²-2yz+x²+z²-2xz=m²+n²+(m+n)²

2x²+2y²+2z²-2xy-2yz-2xz=m²+n²+m²+n²+2mn

x²+y²+z²-xy-yz-xz=m²+n²+mn

平方号好难打啊，终于出来了，嘿嘿

Answer 2

x²+y²+z²-xy-yz-xz=x(x-y)+y(y-z)+z(z-x)
又x-y=m,y-z=n
所以x-z=m+n
代入得右边=(m+n+z)m+(n+z)n+z(-m-n）=m^2+nm+zm+n^2+zn-zn-zm
=m^2+nm+n^2
故x²+y²+z²-xy-yz-xz=m^2+nm+n^2

Answer 3

x-y=m
y-z=n
相加
x-z=m+n

原式=(2x²+2y²+2z²-2xy-2yz-2xz)/2
=[(x²-2xy+y²)+(y²-2yz+z²)+(z²-2xz+x²)]/2
=[(x-y)²+(y-z)²+(x-z)²]/2
=(m²+n²+m²+2mn+n²)/2
=m²+mn+n²

Answer 4

x-y=m,y-z=n
x-z=m+n

x²+y²+z²-xy-yz-xz
=[(x-y)^2+(y-z)^2+(x-z)^2]/2
=[m^2+n^2+(m+n)^2]/2
=m^2+n^2+mn