Question

已知a=(2+根号3)^-1,b=(2-根号3)^-1,求(a+1)^-2+(b+1)^-2的值

Answer 1

a=1/(2+√3)=2-√3
b=1/(2-√3)=2+√3
ab=4-3=1
a+b=4
则a²+b²
=(a+b)²-2ab
=14

原式=1/(a+1)²+1/(b+1)²
=(a²+2a+1+b²+2b+1)/[(a+1)(b+1)]²
=[(a²+b²)+2(a+b)+2]/[ab+(a+b)+1]²
=(14+8+2)/(1+4+1)²
=2/3

Answer 2

a=(2+根号3)^-1=2-根号3,b=(2-根号3)^-1=2+根号3
(a+1)^-2+(b+1)^-2 = (3+根号3)^-2+(3-根号3)^-2=1/(12+6根号3) + 1/(12-6根号3)=(12+6根号3+12-6根号3)/((12+6根号3)(12-6根号3))=24/(144-36*3)=24/36=2/3

Answer 3

解：a=(2+根号3)^-1=2-根号3 b=(2-根号3)^-1=2+根号3 ab=1
原式=(2-根号3)/4

Answer 4

已知a=(2+根号3)^-1=2-根号3
,b=(2-根号3)^-1=2+根号3
(a+1)^-2+(b+1)^-2
=1/(3-根号3)^2+1/(3+根号3)^2
=1/(12-6根号3)+1/(12+6根号3)
=24/(144-108)
=24/36
=2/3