Câu `1`

`a)`

`A = 9x^2 - 3x - 1`

`=  (9x^2 - 3x + 1/4) - 5/4`

`=  [ (3x)^2 - 2 . 3x . 1/2 + (1/2)^2]   - 5/4`

`=  (3x - 1/2)^2 - 5/4 \ge -5/4 \forall x`

Dấu `=` xảy ra `<=> 3x - 1/2 = 0`

`<=>3x=1/2`

`<=> x= 1/6`

Vậy `\text{Min}_A = -5/4 <=>x=1/6`

`b)`

`B = 5x^2 + 2x - 3`

` = 5 (x^2 + 2/5x + 1/25) - 16/5`

` = 5 (x + 1/5)^2 - 16/5 \ge -16/5 \forall x`

Dấu `=` xảy ra `<=>x+1/5=0<=>x=-1/5`

Vậy `\text{Min}_B = -16/5 <=>x=-1/5`

Câu `2`

`a)`

`M = 3 - 5x - 4x^2`

` = - (4x^2 + 5x + 25/16) + 73/16`

`=  - [ (2x)^2 + 2 . 2x . 5/4 + (5/4)^2]  + 73/16`

`=  - (2x + 5/4)^2 + 73/16 \le 73/16 \forall x`

Dấu `=` xảy ra `<= 2x+5/4=0`

`<=> 2x = -5/4`

`<=> x = -5/8`

Vậy `\text{Max}_M = 73/16<=>x=-5/8`

`b)`

`N = 1-  5x^2 + 3x`

` = -5 (x^2 - 3/5x + 9/100) + 29/20`

` = -5 (x - 3/10)^2 + 29/20 \le 29/20 \forall x`

Dấu `=` xảy ra `<=>x-3/10=0<=>x=3/10`

Vậy `\text{Max}_N = 29/20 <=>x=3/10`

Câu `3`

`P = 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 \ge 3/4 \forall x ; y`

Dấu `=` xảy ra `<=> {(x-1/2=0),(y+3=0):} <=> {(x=1/2),(y=-3):}`

Vậy `\text{Min}_P = 3/4 <=> (x;y) = (1/2 ; -3)`

Đáp án+Giải thích các bước giải:

1.

`a)A=9x^2-3x-1`

`=(3x-1/2)^2-5/4`

Vì `(3x-1/2)^2>=0\ ∀x `

`=>(3x-1/2)^2-5/4>=-5/4 \ ∀x`

`Vậy\ MIN_A =-5/4\ khi`

`3x-1/2=0=>x=1/6`

`b)5x^2+2x-3`

`=5.(x+1/5)^2-16/5`

Vì `5.(x+1/5)^2>=0\ ∀x`

`=>5.(x+1/5)^2-16/5>=-16/5 \ ∀ x`

`Vậy\ MIN_B =-16/5\ khi`

`x+1/5=0=>x=-1/5`

2.

`a)M=3-5x-4x^2`

`=-(2x+5/4)^2-73/16`

Vì `-(2x+5/4)^2<=0\ ∀x`

`=>-(2x+5/4)^2-73/16<= -73/16\ ∀x`

`Vậy \ MAX_M =-73/16\ khi`

`2x+5/4=0=>x=-5/8`

`b)N=1-5x^2+3x`

`=-5(x+3/10)^2+29/20`

Vì `-5(x+3/10)^2<=0\ ∀x`

`=>-5(x+3/10)^2+29/20<=29/20\ ∀x`

`Vậy\ MAX_N =29/20\ khi`

`x+3/10=0=>x=-3/10` 

3.

`P = 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`

Vì `(x-1/2)^2+(y+3)^2>=0\ ∀x,y`

`=>(x-1/2)^2+(y+3)^2+3/4>=3/4\ ∀x,y`

Vậy `MIN_P = 3/4\ khi`

`x-1/2=0` và `y+3=0`

`=>x=1/2` và `y=-3`