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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: bai 3: tim GTLL (GTNN) 1, x^2 – 6x + 11 2, 5x – x^2 3, 5 – 8x -x^2 4, 4x -x^2 +3 5, x^2 – 2x + y^2 +4y + 8 6, x^2 – 4x +y^2 -8y +6

Home/ Toán Lớp 8: bai 3: tim GTLL (GTNN) 1, x^2 – 6x + 11 2, 5x – x^2 3, 5 – 8x -x^2 4, 4x -x^2 +3 5, x^2 – 2x + y^2 +4y + 8 6, x^2 – 4x +y^2 -8y +6

Toán Lớp 8: bai 3: tim GTLL (GTNN) 1, x^2 – 6x + 11 2, 5x – x^2 3, 5 – 8x -x^2 4, 4x -x^2 +3 5, x^2 – 2x + y^2 +4y + 8 6, x^2 – 4x +y^2 -8y +6

Tháng Hai 19, 2023 Toán học 2 Comments 0 views

Toán Lớp 8: bai 3: tim GTLL (GTNN)
1, x^2 – 6x + 11
2, 5x – x^2
3, 5 – 8x -x^2
4, 4x -x^2 +3
5, x^2 – 2x + y^2 +4y + 8
6, x^2 – 4x +y^2 -8y +6

Comments ( 2 )

  1. lanngocha
    lanngocha
    19 Tháng Hai, 2023 at 4:17 chiều
    Reply
    ~rai~
    \(1)x^2-6x+11\\\quad\quad=(x^2-6x+9)+2\\\quad\quad=(x-3)^2+2\\\text{Ta có:}(x-3)^2\ge 0\quad\forall x\in\mathbb{R}\\\Leftrightarrow (x-3)^2+2\ge 2\quad\forall x\in\mathbb{R}\\\text{Dấu “=” xảy ra}\Leftrightarrow (x-3)^2=0\Leftrightarrow x=3.\\2)5x-x^2\\=\dfrac{25}{4}-\left(x^2-5x+\dfrac{25}{4}\right)\\=\dfrac{25}{4}-\left(x-\dfrac{5}{2}\right)^2\\\text{Ta có:}\left(x-\dfrac{5}{2}\right)^2\ge 0\quad\forall x\in\mathbb{R}\\\Leftrightarrow -\left(x-\dfrac{5}{2}\right)^2\le 0\quad\forall x\in\mathbb{R}\\\Leftrightarrow \dfrac{25}{4}-\left(x-\dfrac{5}{2}\right)^2\le \dfrac{25}{4}\quad\forall x\in\mathbb{R}\\\text{Dấu “=” xảy ra}\Leftrightarrow \left(x-\dfrac{5}{2}\right)^2=0\\\Leftrightarrow x=\dfrac{5}{2}.\\3)5-8x-x^2\\=21-(x^2+8x+16)\\=21-(x+4)^2\\\text{Ta có:}(x+4)^2\ge 0\quad\forall x\in\mathbb{R}\\\Leftrightarrow -(x+4)^2\le 0\quad\forall x\in\mathbb{R}\\\Leftrightarrow 21-(x+4)^2\le 21\quad\forall x\in\mathbb{R}\\\text{Dấu “=” xảy ra}\Leftrightarrow (x+4)^2=0\Leftrightarrow x=-4.\\4)4x-x^2+3\\=7-(x^2-4x+4)\\=7-(x-2)^2\\\text{Ta có:}(x-2)^2\ge 0\quad\forall x\in\mathbb{R}\\\Leftrightarrow -(x-2)^2\le 0\quad\forall x\in\mathbb{R}\\\Leftrightarrow 7-(x-2)^2\le 7\quad\forall x\in\mathbb{R}\\\text{Dấu “=” xảy ra}\Leftrightarrow (x-2)^2=0\Leftrightarrow x=2.\\5)x^2-2x+y^2+4y+8\\=(x^2-2x+1)+(y^2+4y+4)+3\\=(x-1)^2+(y+2)^2+3\\\text{Ta có:}(x-1)^2\ge 0;(y+2)^2\ge 0\quad\forall x;y\in\mathbb{R}.\\\Leftrightarrow (x-1)^2+(y+2)^2\ge 0\quad\forall x;y\in\mathbb{R}\\\Leftrightarrow (x-1)^2+(y+2)^2+3\ge 3\quad\forall x;y\in\mathbb{R}.\\\text{Dấu “=” xảy ra}\Leftrightarrow \begin{cases}(x-1)^2=0\\(y+2)^2=0\end{cases}\Leftrightarrow \begin{cases}x=1\\y=-2.\end{cases}\\6)x^2-4x+y^2-8y+6\\=(x^2-4x+4)+(y^2-8y+16)-14\\=(x-2)^2+(y-4)^2-14\\\text{Ta có:}(x-2)^2\ge 0;(y-4)^2\ge 0\quad\forall x;y\in\mathbb{R}\\\Leftrightarrow (x-2)^2+(y-4)^2\ge 0\quad\forall x;y\in\mathbb{R}\\\Leftrightarrow (x-2)^2+(y-4)^2-14\ge -14\quad\forall x;y\in\mathbb{R}\\\text{Dấu “=” xảy ra}\Leftrightarrow \begin{cases}(x-2)^2=0\\(y-4)^2=0\end{cases}\Leftrightarrow \begin{cases}x=2\\y=4.\end{cases}\)

  2. lantrungbach
    lantrungbach
    19 Tháng Hai, 2023 at 4:16 chiều
    Reply
    *** Lời giải chi tiết ***
    1)
    x^{2}-6x+11
    =(x^{2}-6x+9)+2
    =(x-3)^{2}+2>=2 với mọi x
    Dấu = xảy ra <=>x-3=0<=>x=3
    Vậy min=2<=>x=3
    2)
    5x-x^{2}
    =-(x^{2}-5x)
    =-(x^{2}-5x+(25)/(4)-(25)/(4))
    =-(x-(5)/(2))^{2}+(25)/(4)≤(25)/(4) với mọi x
    Dấu = xảy ra <=>x-(5)/(2)=0<=>x=(5)/(2)
    Vậy max=(25)/(4)<=>x=(5)/(2)
    3)
    5-8x-x^{2}
    =-(x^{2}+8x-5)
    =-(x^{2}+8x+16-21)
    =-(x+4)^{2}+21≤21 với mọi x
    Dấu = xảy ra <=>x+4=0<=>x=-4
    Vậy max=21<=>x=-4
    4)
    4x-x^{2}+3
    =-(x^{2}-4x-3)
    =-(x^{2}-4x+4-7)
    =-(x-2)^{2}+7≤7 với mọi x
    Dấu = xảy ra <=>x-2=0<=>x=2
    Vậy max=7<=>x=2
    5)
    x^{2}-2x+y^{2}+4y+8
    =(x^{2}-2x+1)+(y^{2}+4y+4)+3
    =(x-1)^{2}+(y+2)^{2}+3≥3 với mọi x;y
    Dấu = xảy ra <=>x=1;y=-2
    Vậy min=3<=>x=1;y=-2
    6)
    x^{2}-4x+y^{2}-8y+6
    =(x^{2}-4x+4)+(y^{2}-8y+16)-14
    =(x-2)^{2}+(y-4)^{2}-14≥ -14 với mọi x;y
    Dấu = xảy ra <=>x=2;y=4
    Vậy min=-14<=>x=2;y=4

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222-9+11+12:2*14+14 = ? ( )

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