要如何求代数式1/x+2/(1-x)的最大值与最小值,要如何求出值域呢？
<p>问题：要如何求代数式1/x+2/(1-x)的最大值与最小值,要如何求出值域呢？<p>答案：↓↓↓<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">董世民的回答：<div class="content-b">网友采纳 x≠0;1y=(1-x+2x)/x(1-x)=(x+1)/(x-x^2)=>y(x-x^2)=x+1=>-yx^2+yx-x-1=0=>yx^2-(y-1)x+1=0;x实数=>判别式[-(y-1)]^2-4y≥0=>y^2-6y+1≥0=>y^2-6y≥-1=>y^2-6y+9≥8=>(y-3)^2≥8=>|y-3|≥2√2=>y-3≥2...
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