x^2-4ax+4a^2+a+1/(a-1）≤0 的解集为空集即(x-2a)^2+a+1/(a-1)≤0的解集为空集那么a+1/(a-1)>0通分(a^2-a+1)/(a-1)>0因为a^2-a+1=(a-1/2)^2+3/4≥3/4>0∴a-1>0a>1判别式法Δ=16a^2-4[4a^2+a+1/(a-1)]0结果一样...