Question

已知a-b=根号3+根号2，b-c=根号3-根号2，求a平方+b平方+c平方-ab-bc-ca

此为二次根式内容

Answer 1

a-b=根号3+根号2，b-c=根号3-根号2，求a^2+b^2+c^2-ab-bc-ca
由上条件可知
a-c=2倍根号3
(a-b)^2=5+2倍根号6
（b-c）^2=5-2倍根号6
(a-c)^2=12
a^2+b^2+c^2-ab-bc-ca
=[(a-b)^2+(a-c)^2+(b-c)^2]/2
=(5+2倍根号6+5-2倍根号6+12）/2
=11

Answer 2

a^2+b^2+c^2-ab-bc-ca
=(2a^2+2b^2+2c^2-2ab-2bc-2ca)/2
=[(a^2-2ab+b^2)+(a^2-2ac+c^2)+(b^2-2bc+c^2)]/2
=[(a-b)^2+(b-c)^2+(a-c)^2]/2
a-b=√3+√2，b-c=√3-√2
a-c=2√3
a^2+b^2+c^2-ab-bc-ca
=[(a-b)^2+(b-c)^2+(a-c)^2]/2
=11

Answer 3

a-b=√3+√2
b-c=√3-√2
a-c=(a-b)+(b-c)=(√3+√2)+(√3-√2)=2√3

a²+b²+c²-ab-bc-ca
=[(a²-2ab+b²)+(b²-2bc+c²)+(c²-2ca+a²)]
=[(a-b)²+(b-c)²+(c-a)²]/2
=[(√3+√2)²+(√3-√2)²+(2√3)²]/2
=[3+2+2√6+3+2-2√6+12]/2
=22/2
=11

Answer 4

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