1/
`x^2-x-6<0`
`⇔x^2-3x+2x-6<0`
`⇔(x-3)(x+2)<0`
`⇔`\(\left[ \begin{array}{l}\left \{ {{x-3>0} \atop {x+1<0}} \right.\\\left \{ {{x-3<0} \atop {x+1>0}} \right.\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}\left \{ {{x>3} \atop {x<-1}} \right.\\\left \{ {{x<3} \atop {x>-1}} \right.\end{array} \right.\)
`⇔-1<x<3`
2/
*) Ta sẽ chứng minh : `(2x^2+5x+2)/(x^2+1)≤9/2`
`⇔(2x^2+5x+2)/(x^2+1)-9/2≤0`
`⇔(4x^2+10x+4-9x^2-9)/[2(x^2+1)]≤0`
`⇔(-5x^2+10x-5)/[2(x^2+1)]≤0`
`⇔-(x^2-2x+1)/[2(x^2+1)]≤0`
`⇔-(x-1)^2/[2(x^2+1)]≤0`
`⇔(x-1)^2/[2(x^2+1)]≥0`
Ta có:
`x^2+1≥0∀x`
`⇔2(x^2+1)≥0∀x`
mà `(x-1)^2≥0∀x`
`⇒(x-1)^2/[2(x^2+1)]≥0∀x`
`⇒(2x^2+5x+2)/(x^2+1)≤9/2∀x`
`⇒`GTLN của `A` là `9/2` đạt khi `x-1=0` hay `x=1`
*) Ta sẽ chứng minh : `(2x^2+5x+2)/(x^2+1)≥-1/2`
`⇔(2x^2+5x+2)/(x^2+1)+1/2≥0`
`⇔(4x^2+10x+4+x^2+1)/[2(x^2+1)]≥0`
`⇔(5x^2+10x+5)/[2(x^2+1)]≥0`
`⇔(x^2+2x+1)/[2(x^2+1)]≥0`
`⇔(x+1)^2/[2(x^2+1)]≥0`
Ta có:
`x^2+1≥0∀x`
`⇔2(x^2+1)≥0∀x`
mà `(x+1)^2≥0∀x`
`⇒(x+1)^2/[2(x^2+1)]≥0∀x`
`⇒(2x^2+5x+2)/(x^2+1)≥-1/2`
`⇒`GTNN của `A` là `-1/2` đạt khi `x+1=0` hay `x=-1`
