当a=2-根号3,b=根号6-根号2时 求代数式(a-b+"a-b分之4ab")(a+b-"a+b分之4ab")

[a-b+4ab/(a-b)]*[a+b-4ab/(a+b)]
=[(a-b)²/(a-b)+4ab/(a-b)]*[(a+b)²/(a+b)-4ab/(a+b) ]
=(a²+b²-2ab+4ab)/(a-b)*(a²+b²+2ab-4ab)/(a+b)
=(a²+b²+2ab)(a²+b²-2ab)/[(a-b)(a+b)]
=(a+b)²(a-b)²/[(a-b)(a+b)]
=(a+b)(a-b)
=a²-b²
=(2﹣√3)²-(√6-√2)²
=7-4√3-(8-4√3)
=7-4√3-8+4√3
=-1

原式=(a²-2ab+b²-4ab)/(a-b)*(a²+2ab+b²-4ab)/(a+b)
=(a+b)²/(a-b)*(a-b)²/(a+b)
=(a+b)(a-b)
=a²-b²
=(7-4√3)-(8-4√3)
=-1是否介意再帮我解一道题 如图在Rt三角形ABC中，...