已知a-b=3,b-c=2求a^2+b^2+c^2-zb-bc-ac的值

来源：百度知道 编辑：UC知道 时间：2024/09/28 10:44:41

a-b=3,b-c=2
a-c=5
a^2+b^2+c^2-zb-bc-ac
=1/2(2a^2+2b^2+2c^2-2cb-2bc-2ac)
=1/2[(a-b)^2+(a-c)^2+(b-c)^2]
=1/2(3^2+2^2+5^2)
=1/2*38
=19

a-b=3
b-c=2
相加
a-c=5

a^2+b^2+c^2-zb-bc-ac
=(2a^2+2b^2+2c^2-2ab-2bc-2ca)/2
=[(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)]/2
=[(a-b)^2+(b-c)^2+(c-a)^2]/2
=(9+4+25)/2
=19

a-b=3,b-c=2
a-c=5
a^2+b^2+c^2-zb-bc-ac
=1/2(2a^2+2b^2+2c^2-2cb-2bc-2ac)
=1/2[(a-b)^2+(a-c)^2+(b-c)^2]
=1/2(3^2+2^2+5^2)
=1/2*38
=19

(z应该为c)

是a^2+b^2+c^2-ab-bc-ac吧
a-c=(a-b)+(b-c)=3+2=5
a^2+b^2+c^2-ab-bc-ac
=(2a^2+2b^2+2c^2-2ab-2bc-2ac)/2
=[(a^2-2ab+b^2)+(b^2-2bc+c^2)+(a^2-2ac+c^2)]/2
=[(a-b)^2+(b-c)^2+(a-c)^2]/2
=(3^2+2^2+5^2)/2
=38/2
=19

a^2+b^2+c^2-ab-bc-ac
=1/2(2a^2+2b^2+2c^2-2ab-2bc-2ac)
=1/2[(a-b)^2+(b-c)^2+(a-c)^2]
=1/2(3^2+2^2+5^2)
=19

设a^2+b^2+c^2-ab-bc-ac=X