Đáp án:

 

`a)`

` A  =x^2- 9x +10 = (x^2 - 9x + 81/4) -41/4 = (x-9/2)^2 -41/4`

Ta có ` (x-9/2)^2 \ge 0 => (x-9/2)^2 -41/4 \ge -41/4`

` => A_{min} = -41/4`

Dấu `=` xảy ra khi ` x-9/2 = 0=> x =9/2`

`b)`

` 9x -3x^2 = -3(x^2 -3x + 9/4) +27/4 = -3* (x-3/2)^2 + 27/4`

Ta có ` (x-3/2)^2 \ge 0 => -3* (x-3/2)^2 \le 0 => B \le 27/4`

Dấu ` =` xảy ra khi ` x -3/2 = 0 => x = 3/2`

`c)`

` x^2 +y^2 -x + 6y+10`

` = (x^2 -x +1/4) + (y^2 +6y + 9) + 3/4` 

 ` =(x-1/2)^2 + (y+3)^2 +3/4`

Ta có ` (x-1/2)^2 \ge 0 ; (y+3)^2 \ge 0`

` => C \ge 3/4`

` => C_{min} = 3/4`

Dấu `=` xảy ra khi ` x-1/2 = 0 => x= 1/2`

 

a) $A = x^2-9x+10$

$ = \bigg(x-\dfrac{9}{2}\bigg)^2  - \dfrac{41}{4} ≥ \dfrac{-41}{4}$

Dấu "=" xảy ra $⇔x=\dfrac{9}{2}$

b) $B = 9x-3x^2$

$ \to -B = 3x^2-9x$

$ = 3.(x^2-3x)$

$ = 3.\bigg(x-\dfrac{3}{2}\bigg)^2-\dfrac{27}{4} ≥ \dfrac{-27}{4}$

$\to B ≤ \dfrac{27}{4}$

Dấu "=" xảy ra $⇔x=\dfrac{3}{2}$

c) $C  = x^2+y^2-x+6y+10$

$ = \bigg(x-\dfrac{1}{2}\bigg)^2 + (y+3)^2 + \dfrac{3}{4} ≥ \dfrac{3}{4}$

Dấu "=" xảy ra khi $x=\dfrac{1}{2}l y=  -3$