已知A-B=根号3+根号2;B-C=根号3-根号2求a2+b2+c2-ab-bc-ca
第1个回答 2009-09-09
a-b=√3+√2
b-c=√3-√2
相加
a-c=2√3
a²+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)²]/2
=[(√3+√2)²+(√3-√2)²+(2√3)²]/2
=(3+2√6+2+3-2√6+2+12)/2
=11本回答被提问者采纳
b-c=√3-√2
相加
a-c=2√3
a²+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)²]/2
=[(√3+√2)²+(√3-√2)²+(2√3)²]/2
=(3+2√6+2+3-2√6+2+12)/2
=11本回答被提问者采纳
第2个回答 2009-09-09
∵(A-B)+(B-C)=A-C=2根号3
(A-B)^2+(A-C)^2+(B-C)^2
=A^2+B^2-2AB+A^2+C^2-2AC+B^2+C^2-2BC
=2A^2+2B^2+2C^2-2AB-2AC-2BC
=2(A^2+B^2+C^2-AB-AC-BC)
=(根号3+根号2)^2+(根号3-根号2)^2+(2根号3)^2
=3+2+2根号6+3+2-2根号6+12
=22
∴A^2+B^2+C^2-AB-AC-BC=11
(A-B)^2+(A-C)^2+(B-C)^2
=A^2+B^2-2AB+A^2+C^2-2AC+B^2+C^2-2BC
=2A^2+2B^2+2C^2-2AB-2AC-2BC
=2(A^2+B^2+C^2-AB-AC-BC)
=(根号3+根号2)^2+(根号3-根号2)^2+(2根号3)^2
=3+2+2根号6+3+2-2根号6+12
=22
∴A^2+B^2+C^2-AB-AC-BC=11
第3个回答 2009-09-09
a-b =√3 + √2 ①
b-c =√3 - √2 ②
所以 (a-b)²= a² + b² - 2ab = (√3 + √2)² = 3 + 2 + 2√6 = 5 + 2√6 ④
所以 (b-c)²= b² + c² - 2bc = (√3 - √2)² = 3 + 2 - 2√6 = 5 - 2√6 ⑤
所以 (a-c)²= a² + c² - 2ac = (① + ②)² = (2√3)² = 12 ⑥
a²+b²+c² - ab - bc - ca = ( ④ + ⑤ + ⑥ ) / 2 = (5 + 2√6 + 5 - 2√6 + 12)/2 = 22/2 = 11
b-c =√3 - √2 ②
所以 (a-b)²= a² + b² - 2ab = (√3 + √2)² = 3 + 2 + 2√6 = 5 + 2√6 ④
所以 (b-c)²= b² + c² - 2bc = (√3 - √2)² = 3 + 2 - 2√6 = 5 - 2√6 ⑤
所以 (a-c)²= a² + c² - 2ac = (① + ②)² = (2√3)² = 12 ⑥
a²+b²+c² - ab - bc - ca = ( ④ + ⑤ + ⑥ ) / 2 = (5 + 2√6 + 5 - 2√6 + 12)/2 = 22/2 = 11
第4个回答 2020-07-08
a-b=√3+√2
b-c=√3-√2
相加
a-c=2√3
a²+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)²]/2
=[(√3+√2)²+(√3-√2)²+(2√3)²]/2
=(3+2√6+2+3-2√6+2+12)/2
=11
b-c=√3-√2
相加
a-c=2√3
a²+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)²]/2
=[(√3+√2)²+(√3-√2)²+(2√3)²]/2
=(3+2√6+2+3-2√6+2+12)/2
=11
第5个回答 2009-09-09
a-c=2√3
a^2+b^2+c^2-ab-bc-ca
=1/2(2a^2+2b^2+2c^2-2ab-2ac-2bc)
=1/2[(a-b)^2+(b-c)^2+(c-a)^2]
=1/2[(√3+√2)^2+(√3-√2)^2+(2√3)^2]
=1/2(5+2√6+5-2√6+12)
=1/2*22
=11
a^2+b^2+c^2-ab-bc-ca
=1/2(2a^2+2b^2+2c^2-2ab-2ac-2bc)
=1/2[(a-b)^2+(b-c)^2+(c-a)^2]
=1/2[(√3+√2)^2+(√3-√2)^2+(2√3)^2]
=1/2(5+2√6+5-2√6+12)
=1/2*22
=11