求值域y=x+(x^2-3x+2)^1/2
<p>问题：求值域y=x+(x^2-3x+2)^1/2<p>答案：↓↓↓<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">李东宁的回答：<div class="content-b">网友采纳 由y=x+√(x²-3x+2)得√(x²-3x+2)=y-x≥0两边平方,得(2y-3)x=y²-2,从而,y≠3/2,且x=(y²-2)/(2y-3).由y-x=y-(y²-2)/(2y-3)≥0,得(y²-3y+2)/(2y-3)≥0,1≤y〈3/2或y≥2.当y≥2时,由x=...
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