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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")
人气:227 ℃ 时间:2019-08-19 06:43:39
解答

[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
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