Question

高一指数函数题

设x＝a﹢3a^(1/3)×b^(﹣2/3)，y＝b^﹣1﹢3a^(2/3)×b^(﹣1/3)。若a^(2/3)﹢b^(﹣2/3)＝1，求(x﹢y)^(2/3)﹢(x﹣y)^(2/3)的值

Answer 1

∵x＝a﹢3a^(1/3)×b^(﹣2/3)，
y＝b^﹣1﹢3a^(2/3)×b^(﹣1/3)。
∴x+y=a+b^(-1)+3a^(1/3)×b^(﹣2/3)+3a^(2/3)×b^(﹣1/3)
=[a^(1/3)]³+3a^(2/3)×b^(﹣1/3)+3a^(1/3)×b^(﹣2/3)++[b^(-1/3)]³
=[a^(1/3)+b(-1/3)]³
x-y=a-b^(-1)-3a^(1/3)×b^(﹣2/3)+3a^(2/3)×b^(﹣1/3)
=[a^(1/3)]³+3a^(2/3)×b^(﹣1/3)+3a^(1/3)×b^(﹣2/3)++[b^(-1/3)]³
=[a^(1/3)-b(-1/3)]³
又a^(2/3)﹢b^(﹣2/3）=1
∴ (x﹢y)^(2/3)﹢(x﹣y)^(2/3)
=[a^(1/3)+b(-1/3)]²+[a^(1/3)-b(-1/3)]²
=a^(2/3)﹢b^(﹣2/3)+2a^(1/3)+b(-1/3)+a^(2/3)﹢b^(﹣2/3)-2a^(1/3)+b(-1/3)
=2[a^(2/3)﹢b^(﹣2/3)]＝2
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