Tìm giá trị nhỏ nhất: b) Q = 2x ² – 6x c) M = x ² + y ² – x + 6y + 10 d) N = x ² + 5x + 7 07/07/2021 Bởi Margaret Tìm giá trị nhỏ nhất: b) Q = 2x ² – 6x c) M = x ² + y ² – x + 6y + 10 d) N = x ² + 5x + 7
`Q=2x^2-6x` `=2(x^2-3x)` `=2(x^2-2.x*3/2+9/4)-9/2` `=2(x-3/2)^2-9/2>=-9/2` Dấu “=” xảy ra khi `x=3/2` `M=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>=3/4` Dấu “=” xảy ra khi `x=1/2,y=-3` `N=x^2+5x+7` `=x^2+5x+25/4+3/4` `=(x+5/2)^2+3/4>=3/4` Dấu “=” xảy ra khi `x=-5/2`. Bình luận
Giải a) Ta có: `Q=2x^2 – 6x` `Q=2.(x^2 – 3x)` `Q= 2.(x^2 – 2. 3/2 x + 9/4) – 9/2` `Q=2.(x-3/2)^2 – 9/2` Vì `(x-3/2)^2 ≥ 0` `⇒ 2.(x-3/2)^2 ≥ 0 ` `⇒ 2. (x-3/2) – 9/2 ≥ -9/2` Dấu “=” xảy ra ⇔ `2. (x-3/2) – 9/2 = -9/2` ⇔ `2.(x-3/2) = 0` ⇔ `x-3/2 = 0` ⇔ `x= 3/2` Vậy `GTNN` của `Q = -9/2` khi `x=3/2` b) Ta có: `M = x ² + y ² – x + 6y + 10` `M=(x^2 – x) + (y^2 + 6y + 9) + 1` `M=(x^2 – 2. 1/2x + 1/4) + (y+3)^2 + 3/4` `M=(x-1/2)^2 + (y+3)^2 + 3/4` Vì `(x-1/2)^2 ≥ 0` `(y+3)^2 ≥ 0` ⇒ `(x-1/2)^2 + (y+3)^2 ≥ 0` ⇒ `(x-1/2)^2 + (y+3)^2 + 3/4 ≥ 3/4` Dấu “=” xảy ra ⇔ `(x-1/2)^2 + (y+3)^2 = 0` ⇔ $\left \{ {{x-1/2=0} \atop {y+3=0}} \right.$ ⇔ $\left \{ {{x=1/2} \atop {y=-3}} \right.$ Vậy …. c) Ta có: `N = x ² + 5x + 7` `N=(x^2 + 2. 5/2x + 25/4) + 3/4` `N = (x+5/2)^2 + 3/4` Vì `(x+5/2)^2 ≥ 0` ⇒ `(x+5/2)^2 + 3/4 ≥ 3/4` Dấu “=” xảy ra ⇔ `(x+5/2)^2=0` `⇔ x+5/2=0` `⇔ x=-5/2` Vậy…. Bình luận
`Q=2x^2-6x`
`=2(x^2-3x)`
`=2(x^2-2.x*3/2+9/4)-9/2`
`=2(x-3/2)^2-9/2>=-9/2`
Dấu “=” xảy ra khi `x=3/2`
`M=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>=3/4`
Dấu “=” xảy ra khi `x=1/2,y=-3`
`N=x^2+5x+7`
`=x^2+5x+25/4+3/4`
`=(x+5/2)^2+3/4>=3/4`
Dấu “=” xảy ra khi `x=-5/2`.
Giải
a) Ta có: `Q=2x^2 – 6x`
`Q=2.(x^2 – 3x)`
`Q= 2.(x^2 – 2. 3/2 x + 9/4) – 9/2`
`Q=2.(x-3/2)^2 – 9/2`
Vì `(x-3/2)^2 ≥ 0`
`⇒ 2.(x-3/2)^2 ≥ 0 `
`⇒ 2. (x-3/2) – 9/2 ≥ -9/2`
Dấu “=” xảy ra ⇔ `2. (x-3/2) – 9/2 = -9/2`
⇔ `2.(x-3/2) = 0`
⇔ `x-3/2 = 0`
⇔ `x= 3/2`
Vậy `GTNN` của `Q = -9/2` khi `x=3/2`
b) Ta có: `M = x ² + y ² – x + 6y + 10`
`M=(x^2 – x) + (y^2 + 6y + 9) + 1`
`M=(x^2 – 2. 1/2x + 1/4) + (y+3)^2 + 3/4`
`M=(x-1/2)^2 + (y+3)^2 + 3/4`
Vì `(x-1/2)^2 ≥ 0`
`(y+3)^2 ≥ 0`
⇒ `(x-1/2)^2 + (y+3)^2 ≥ 0`
⇒ `(x-1/2)^2 + (y+3)^2 + 3/4 ≥ 3/4`
Dấu “=” xảy ra ⇔ `(x-1/2)^2 + (y+3)^2 = 0`
⇔ $\left \{ {{x-1/2=0} \atop {y+3=0}} \right.$
⇔ $\left \{ {{x=1/2} \atop {y=-3}} \right.$
Vậy ….
c) Ta có: `N = x ² + 5x + 7`
`N=(x^2 + 2. 5/2x + 25/4) + 3/4`
`N = (x+5/2)^2 + 3/4`
Vì `(x+5/2)^2 ≥ 0`
⇒ `(x+5/2)^2 + 3/4 ≥ 3/4`
Dấu “=” xảy ra ⇔ `(x+5/2)^2=0`
`⇔ x+5/2=0`
`⇔ x=-5/2`
Vậy….