已知:a-b=8,a-c=6.求:a05+b05+c05-ab-bc-ac的值
问题描述:
已知:a-b=8,a-c=6.求:a05+b05+c05-ab-bc-ac的值
答
a-b=8,a-c=6相减b-c=-2a²+b²+c²-ab-bc-ac=(2a²+2b²+2c²-2ab-2bc-2ac)/2=[(a²-2ab+b²)+(b²-2bc+c²)+(c²-2ac+a²)]/2=[(a-b)²+(b-c)²+(a-c)...