[(√a-√b)³+2a²÷√a+b√b]/3(a√a+b√b)+(√ab-b)/(a-b)
={[(√a)³-(√b)³-3a√b+3b√a]+2a√a*√a÷√a+b√b}/3[(√a)³+(√b)³]+√b(√a-√b)/(a-b)
=[a√a-b√b-3a√b+3b√a+2a√a+b√b]/3[(√a+√b)(√a²-√a√b+√b²)]+√b(√a-√b)/(√a-√b)(√a+√b)
=(3a√a-3a√b+3b√a)/3[(√a+√b)(a-√a√b+b)]+√b/(√a+√b)
=3√a(a-√a√b+b)/3[(√a+√b)(a-√a√b+b)]+√b/(√a+√b)
=√a/(√a+√b)+√b/(√a+√b)
=(√a+√b)/(√a+√b)
=1
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