(2倍根号3+3倍根号2-根号6)(2倍根号3-3倍根号2+根号6)

如题所述

(2倍根号3+3倍根号2-根号6)(2倍根号3-3倍根号2+根号6)

=[2倍根号3+(3倍根号2-根号6)][2倍根号3-(3倍根号2-根号6)]

=(2根号3)^2-(3根号2-根号6)^2

=12-(18+6-12根号3)
=-12+12根号3
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第1个回答  2009-09-10
8.784609691
第2个回答  2019-05-30
(2倍根号3-3倍根号2-根号6)(2倍根号3+3倍根号2-根号6)
=(2倍根号3-根号6-3倍根号2)(2倍根号3-根号6+3倍根号2)
=(2倍根号3-根号6)^2-(3倍根号2)^2
=12-12倍根号2+6-18
=-12倍根号2
根号27+[2+根号3]分之1-[根号3-1分之根号3]o次幂
=根号27+[2+根号3]分之1-1
=3倍根号3+2-根号3-1
=2倍根号3+1
根号12+[2-根号3]分之1-[2+根号3]乘根号3
=2倍根号3+2+根号3-(2倍根号3+3)
=根号3-1
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