已知a-b=1/(2-√3),b-c=1/(2+√3),求a^2+b^2+c^2-ab-bc-ca的值 5
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解：因为a-b=1/(2-√3), b-c=1/(2+√3)
所以a-b+b-c=1/(2-√3)+1/(2+√3)
a-c=(2+√3)/[(2-√3)(2+√3)]+(2-√3)/[(2-√3)(2+√3)]
=(2+√3)/(4-3)+(2-√3)/(4-3)
=2+√3+2-√3
=4
因为
a²+b²+c²-ab-bc-ca
=(2a²+2b²+2c²-2ab-2bc-2ca)/2
=[(a²-2ab+b²)+(b²-2bc+c²)+(a²-2ca+c²)]/2
=[(a-b)²+(b-c)²+(a-c)²]/2
=[1/(2-√3)²+1/(2+√3)²+4²]/2
=[1/(7-4√3)+1/(7+4√3)+16]/2
=﹛(7+4√3)/[(7-4√3)(7+4√3)]+(7-4√3)/[(7-4√3)(7+4√3)]+16﹜/2
=[(7+4√3)/(49-48)+(7-4√3)/(49-48)+16]/2
=(7+4√3+7-4√3+16)/2
=15
所以a-b+b-c=1/(2-√3)+1/(2+√3)
a-c=(2+√3)/[(2-√3)(2+√3)]+(2-√3)/[(2-√3)(2+√3)]
=(2+√3)/(4-3)+(2-√3)/(4-3)
=2+√3+2-√3
=4
因为
a²+b²+c²-ab-bc-ca
=(2a²+2b²+2c²-2ab-2bc-2ca)/2
=[(a²-2ab+b²)+(b²-2bc+c²)+(a²-2ca+c²)]/2
=[(a-b)²+(b-c)²+(a-c)²]/2
=[1/(2-√3)²+1/(2+√3)²+4²]/2
=[1/(7-4√3)+1/(7+4√3)+16]/2
=﹛(7+4√3)/[(7-4√3)(7+4√3)]+(7-4√3)/[(7-4√3)(7+4√3)]+16﹜/2
=[(7+4√3)/(49-48)+(7-4√3)/(49-48)+16]/2
=(7+4√3+7-4√3+16)/2
=15
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