Tìm GTNN của : `a)\sqrt{2x^2 -4x+5}+1` `b)\sqrt{x(x+1)(x+2)(x+3)+5}`

`a)`

`sqrt(2x^2-4x+5)` `+1`

`=` `sqrt(2x^2-4x+2+3)` `+1`

`=` `sqrt(2(x^2-2x+1)+3)` `+1`

`=` `sqrt(2(x-1)^2+3)` `+1`

Ta có :

`(x-1)^2` `\geq` `0` `∀x`

`⇒` `2(x-1)^2` `\geq` `0` 

`⇒` `2(x-1)^2+3` `\geq` `3` 

`⇒` `sqrt(2(x-1)^2+3)` `\geq` `sqrt3` 

`⇒` `sqrt(2(x-1)^2+3)` `+1` `\geq` `1+sqrt3` `∀x`

`⇒` `A` `\geq` `1+sqrt3` `∀x`

Dấu "=" xảy ra khi `(x-1)^2` `=0`

`⇔` `x=1`

Vậy Min `sqrt(2x^2-4x+5)` `=` `1+sqrt3` khi `x=1`

`b)`

`\sqrt{x(x+1)(x+2)(x+3)+5}`

`=` `sqrt(x(x+3)(x+1)(x+2)+5)`

`=` `sqrt((x^2+3x)(x^2+3x+2)+5)`

Đặt : `t=x^2+3x`

Ta được : 

`sqrt((x^2+3x)(x^2+3x+2)+5)` `=` `t(t+2)+5`

`=` `t^2+2t+1+4`

`=` `(t+1)^2+4`

Ta có :

`(t+1)^2` `\geq` `0` `∀x`

`⇒` `(t+1)^2+4` `\geq` `4` `∀x`

Dấu "=" xảy ra khi `(t+1)=0`

`⇒` `x^2+3x+1` `=` `0`

`⇔` `x^2+3x` `=` `-1`

`⇔` `x=(-3±sqrt5)/2`

Vậy Min `\sqrt{x(x+1)(x+2)(x+3)+5}` `=4` khi `x=(-3±sqrt5)/2`