`E = x^2 - 6x + 3`
` = (x^2 - 6x + 9) -6`
` = (x^2 - 2 . x . 3 + 3^2) - 6`
`= (x-3)^2 - 6`
`\forall x ` ta có :
`(x-3)^2 \ge 0`
`=> (x-3)^2 - 6 \ge -6`
`=> E \ge -6`
Dấu `=` xảy ra `<=> x - 3 =0`
`<=> x=3`
Vậy `\text{Min}_E = -6 <=> x = 3`
``
`F = 5x^2 + 10x + 2015`
` = 5 (x^2 + 2x + 1) + 2010`
` = 5 (x^2 + 2 . x . 1 + 1^2) + 2010`
` = 5 (x+1)^2 + 2010`
`\forall x ` ta có :
`(x+1)^2 \ge 0`
`=> 5(x+1)^2 \ge 0`
`=> 5(x+1)^2 + 2010 \ge 2010`
`=> F \ge 2010`
Dấu `=` xảy ra `<=> x + 1 =0`
`<=> x = -1`
Vậy `\text{Min}_F = 2010 <=> x = -1`
``
`C = 5x^2 + 4y^2 + 10x - 12y + 2031`
` = (5x^2 + 10x + 5 ) + (4y^2 - 12y + 9 ) + 2017`
` = [5 (x^2 + 2x + 1) ] + [ (2y)^2 - 2 . 2y . 3 + 3^2 ] + 2017`
` = 5 (x+1)^2 + (2y-3)^2 + 2017`
`\forall x ; y` ta có :
`5(x+1)^2 \ge 0`
`(2y-3)^2 \ge 0`
`=> 5(x+1)^2 + (2y-3)^2 \ge 0`
`=> 5(x+1)^2 + (2y-3)^2 + 2017 \ge 2017`
`=> C \ge 2017`
Dấu `=` xảy ra `<=>` $\begin{cases} x + 1 =0 \\ 2y - 3=0 \end{cases}$
`<=>` $\begin{cases} x =-1 \\ y = 1,5 \end{cases}$
Vậy `\text{Min}_C = 2017 <=> `$\begin{cases} x =-1 \\ y = 1,5 \end{cases}$
