Tìm GTLN hoặc GTNN D = (x+ 1) (x + 2) (x + 3) (x + 4) + 2011 F = $\frac{3y^{2}}{-25x^2+20xy-5y^2}$

Ta có: `D=(x+1)(x+2)(x+3)(x+4)+2011`

`=[(x+1)(x+4)].[(x+2)(x+3)]+2011`

`=(x^2 +4x+x+4)(x^2 +3x+2x+6)+2011`

`=(x^2+5x+4)(x^2 +5x+6)+2011`

$Đặt:$ `t=x^2+5x+4`

`⇔D=(t-1)(t+1)+2011`

`⇔D=t^2-1+2011`

`⇔D=t^2+2010`

`⇔D=(x^2+5x+5)^2+2010`

Vì `(x^2+5x+5)^2≥0` `∀x`

`⇔D≥2010` `∀x`

Dấu " `=` " xảy ra khi:
        `(x^2+5x+5)^2=0`

`⇔x^2+5x+5=0`

`⇔(x^2 +2.x.5/2 + 25/4)-5/4=0`

`⇔(x+5/2)^2- ( \sqrt{5/4} )^2=0`

`⇔(x+5/2-\sqrt{5/4})(x+5/2+\sqrt{5/4})=0`

`⇔[(x+5/2-\sqrt{5/4}=0),(x+5/2+\sqrt{5/4}=0):}`

`⇔[(x=-(5-\sqrt{5} )/2 ),(x=-(5+\sqrt{5})/2 ):}` 

Vậy, $Min_D$ `=2010` tại `x∈{-(5-\sqrt{5} )/2 , -(5+\sqrt{5})/2)`

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Ta có: `F= (3y^2)/(-25x^2 +20xy - 5y^2)`

`= - (3y^2)/( 25x^2 - 20xy + 4y^2 + y^2)`

`= -(3y^2)/( (5x-2y)^2 + y^2 )`

Vì `(5x-2y)^2 ≥0` `x,y`

`⇒(5x-2y)^2 + y^2≥y^2` `x,y`

`⇒F≥ - ( 3y^2 )/(y^2) = -3`

Vậy, $Min_F$ `=-3` tại `5x=2y⇔x=2/5y`

`#` $Mpro$