Question

Tìm GTNN của biểu thức sau:3x²+5x+4

Answer 1

\(3x^2+5x+4=3\left(x^2+\dfrac{5}{3}x+\dfrac{25}{36}\right)+\dfrac{23}{12}=3\left(x+\dfrac{5}{3}\right)^2+\dfrac{23}{12}\ge\dfrac{23}{12}\forall x\)

Dấu "=" <=> \(x=-\dfrac{5}{3}\)

Answer 2

<=>\(3\left(x^2+\dfrac{5}{3}x+\dfrac{4}{3}\right)< =>3\left(x^2+2.\dfrac{5}{6}x+\dfrac{25}{9}-\dfrac{13}{9}\right)\)

<=>\(3\left(x+\dfrac{5}{3}\right)^2-\dfrac{13}{3}\)

Vay GTNN =\(-\dfrac{13}{3}< =>x=-\dfrac{5}{3}\)

 

Answer 3

3x² + 5x + 4

= 3(x² + 5/3 x + 4/3)

= 3(x² + 2.5/6 x + 25/36 + 23/36)

= 3[(x + 5/6)² + 23/36]

= 3(x + 5/6)² + 23/12

Do (x + 5/6)² >= 0

Suy ra 3(x + 5/6)² >= 0

Suy ra 3(x + 5/6)² + 23/12 >= 23/12

Vậy GTNN của biểu thức đã cho là 23/12 khi x = -5/6

Answer 4

`A=3x^2+5x+4`

`A= 3.( x^2 + 5/6 . x +4/3)`

`A= 3.( x^2 + 2. 5/6 x + 48/36)`

`A= 3.( x^2+ 2 . 5/6 x + 25/36+ 23/36)`

`A=3.(x^2 + 2. 5/6 x + 25/36)+ 23/12`

`A= 3.(x+5/6)^2 + 23/12>=x`

dấu ''`=`'' xảy ra khi `x+5/6=0=>x=-5/6`

vậy \(MinA\)`=23/12<=>x=-5/6`