y=(3x2+2x+1)/(x-1)求值域

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y=(3x2+2x+1)/(x-1)求值域y=(3x2+2x+1)/(x-1)求值域y=(3x2+2x+1)/(x-1)求值域令t=x-1x=t+1y=[3(t^2+2t+1)+2t+2+1]/t=(

y=(3x2+2x+1)/(x-1)求值域
y=(3x2+2x+1)/(x-1)求值域

y=(3x2+2x+1)/(x-1)求值域
令t=x-1
x=t+1
y=[3(t^2+2t+1)+2t+2+1]/t=(3t^2+8t+6)/t=8+3(t+2/t)
由均值不等式:|t+2/t|>=2 √2
所以y>=8+3*2 √2=8+6 √2
或y

设t=x-1, 则x=t+1,采用换元的方法,y=3t+6/t+8,值域是大于等于6根号2+8