非礼李宇春
幼苗
共回答了26个问题采纳率:84.6% 举报
∵ a³(x+y)²×(x-y)≥0,x﹥y﹥0
∴ x-y﹥0,x+y﹥0,a≥0
原式=√[a³(x+y)²(x-y)]
=√[a²(x+y)²×a(x-y)]
=a(x+y)√[a(x-y)]
∵p²/(p-q)≥0,p﹥q﹥0
∴p﹥0,p-q﹥0
原式=√[p²/(p-q)]
=√[p²(p-q)/(p-q)²]
=√[p²/(p-q)²]×√(p-q)
=p/(p-q)×√(p-q)
=[p√(p-q)]/(p-q)
1年前
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