∵ 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)...