Question

x-y＝a y-z＝B 求 X²＋Y²＋Z²-XY-YZ-XZ的值

Answer 1

X-Z=A+B
X²＋Y²＋Z²-XY-YZ-XZ=05（2X²＋2Y²＋2Z²-2XY-2YZ-2XZ）
=0.5[（x-y）^2+(y-z)^2+(X-z)^]
=0.5(a^2+b^2+a^2+b^2+2ab)
=a^2+b^2+ab

Answer 2

x-y=a
y-z=b
相加
x-z=a+b

x²+y²+z²-xy-yz-xz
=(1/2)(2x²+2y²+2z²-2xy-2yz-2zx)
=(1/2)(x²-2xy+y²+y²-2yz+z²+z²-2zx+x²)
=(1/2)[(x-y)²+(y-z)²+(x-z)²]
=(1/2)[a²+b²+(a+b)²]
=(1/2)(2a²+2b²+2ab)
=a²+ab+b²

Answer 3

因为x-y＝a y-z＝B，两式相加得 x-z=a-B
设S=X2＋Y2＋Z2-XY-YZ-XZ,则：
2S=2（X2＋Y2＋Z2-XY-YZ-XZ）
=(x-y)^2+(x-z)^2+(y-z)^2
=a^2+B^2+(a+B)^2
=2a^2+2B^2+2aB
所以，S=a^2+aB+B^2.

Answer 4

x-y=a,y-z=b,则x-z=(x-y)+(y-z)=a+b.
原式=[(x-y)²+(y-z)²+(x-z)²]/2
=[a²+b²+(a+b)²]/2
=a²+ab+b².
请采纳！谢谢！

Answer 5

x-y＝a y-z＝b两式相加得x-z=a+b X²＋Y²＋Z²-XY-YZ-XZ=1/2[(x-y)²+(y-z)²+(z-x)²]=1/2[a²+b²+(a+b)²]=a²+b²+ab