解:a=1/(√3+√2) =(√3-√2)/[(√3+√2)(√3-√2)] =√3-√2a-1/a=-2√2a+1/a=2√3√〔(a-1/a)^2+4〕-√〔(a+1/a)^2+4〕=√(8+4)-√(12+4)=2√3-4
a=1/(√3-√2)=√3+√2; 1/a=√3-√2;a-1/a=2√2; a+1/a=2√3;√〔(a-1/a)^2+4〕-√〔(a+1/a)^2+4〕=√[(2√2)²+4]-√[(2√3)²+4]=√12-√16=2√3-4