(a+b)^2=a^2+2ab+b^2
∵(a-b)^2=a^2-2ab+b^2=(2根号3-1)^2=13-4根号3
∴a^2+b^2=13-4根号3+2·根号3=13-2根号3
即(a+b)^2=13-2根号3+2·根号3=13
∴a+b=±根号13
即(a+b)(a-b)=±根号13·(2根号3-1)
答案:2根号39-根号13或根号13-2根号39
a-b=2√3-1
两边平方
(a-b)²=12-4√3+1
a²-2ab+b²=13-4√3
a²+b²=13-4√3+2ab=13-2√3
(a+b)²
=a²+2ab+b
=13
a+b=±√13
所以原式=±√13(2√3-1)
=2√39-√13或-2√39+√13
a-b=2根号3-1,ab=根号3,则a+b=2根号3+1,所以(a+b)(a-b)=(2根号3+1)*(2根号3-1)=11
(a-b)²=(2√3-1)²=13-4√3
∴(a+b)²-4ab=13-4√3
∴a+b= ±√13
(a+b)(a-b)=±√13(2√3-1)