1、x<1,求(x^2+3x+4)⼀(x-1)的取值范围 2、x>1,求(x^2+1)⼀(x-1)的取值范围 急急急~~~

2025-03-01 19:09:23
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回答1:

解:①原式=[(x-1)²+5(x-1)+8]/(x-1)=5-[(1-x)+8/(1-x)]≤5-2根号8=5-4根号2
②原式=[(x-1)²+2(x-1)+2]/(x-1)=2+[(x-1)+2/(x-1)]≥2+2根号2