y=(3x+1)/(x-2) ==> x=(1+2y)/(y-3),y=3时此式无意义,故y不=3,即函数值域为(-无穷,3)U(3,无穷)。
y=3x+1/x-2 当x>0时y≥2√3-2,当且仅当3x=1/x,即x=√3时取等号 当x<0时y=-[(-3x)+(-1/x)]-2≤-2√3-2,当x=-√3时取等号 所以y∈(-∞,-2-2√3]∪[-2+2√3,+ ∞)
x-2>0即x>2 y=3(x-2)+1/(x-2)+6>6+2√3 x-2<0即x<2 y=3(x-2)+1/(x-2)+6<6-2√3 ∴y∈(-∞,6-2√3)∪(6+2√3,+ ∞)
答：y=3x+2/(x-2)=3(x-2)+2/(x-2)+6>=2*{[3(x-2)*(2/(x-2))]^(1/2)}+6=6+2*6^(1/2)即y>=6+2...详情>>
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