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`
