\(C=16x^2-8x+2024\)

\(\Rightarrow C=16x^2-8x+1+2023\)

\(\Rightarrow C=\left(4x-1\right)^2+2023\ge2023\left(\left(4x-1\right)^2\ge0\right)\)

\(\Rightarrow Min\left(C\right)=2023\)

\(D=-25x^2+50x-2023\)

\(\Rightarrow D=-\left(25x^2-50x+25\right)-1998\)

\(\Rightarrow D=-\left(5x-5\right)^2-1998\le1998\left(-\left(5x-5\right)^2\le0\right)\)

\(\Rightarrow Max\left(D\right)=1998\)

\(B=-x^2+20x+100=-\left(x^2-20x+100\right)+200=-\left(x-10\right)^2+200\le200\left(-\left(x-10\right)^2\le0\right)\)

\(\Rightarrow Max\left(B\right)=200\)

\(E=\left(2x-1\right)^2-\left(3x+2\right)\left(x-5\right)\)

\(\Rightarrow E=4x^2-4x+1-\left(3x^2-13x-10\right)\)

\(\Rightarrow E=4x^2-4x+1-3x^2+13x+10\)

\(\Rightarrow E=x^2+9x+11=x^2+9x+\dfrac{81}{4}-\dfrac{81}{4}+11\)

\(\Rightarrow E=\left(x+\dfrac{9}{2}\right)^2-\dfrac{37}{4}\ge-\dfrac{37}{4}\left(\left(x+\dfrac{9}{2}\right)^2\ge0\right)\)

\(\Rightarrow Min\left(E\right)=-\dfrac{37}{4}\)

\(F=\left(3x-5\right)^2-\left(3x+2\right)\left(4x-1\right)\)

\(\Rightarrow F=9x^2-30x+25-\left(12x^2+3x-2\right)\)

\(\Rightarrow F=-3x^2-33x+27=-3\left(x^2-10x+9\right)\)

\(\Rightarrow F=-3\left(x^2-10x+25\right)+48=-3\left(x-5\right)^2+48\le48\left(-3\left(x-5\right)^2\le0\right)\)

\(\Rightarrow Max\left(F\right)=48\)

Ta có

B   =   4   –   16 x 2   –   8 x     =   5   –   ( 16 x 2   +   8 x   +   1 )   =   5   –   [ ( 4 x ) 2   +   2 . 4 x . 1   +   1 2 ]     =   5   –   ( 4 x   +   1 ) 2

Nhận thấy 4 x   +   1 2 ≥ 0; Ɐx

=> 5 – 4 x   +   1 2 ≤ 5

Dấu “=” xảy ra khi 4 x   +   1 2 = 0 ó x = - 1 4

Đáp án cần chọn là: A

`A=16x^2+8x+5`

`=16x^2+8x+1+4`

`=(4x+1)^2+4>=4`

Dấu "=" xảy ra khi `4x+1=0<=>x=-1/4`

`B=x^2-x`

`=x^2-x+1/4-1/4`

`=(x-1/2)^2-1/4>=-1/4`

Dấu "=" xảy ra khi `x=1/2`

`C=a^2-2a+b^2+6b+2021`

`=a^2-2a+1+b^2+6b+9+2011`

`=(a-1)^2+(b+3)^2+2011>=2011`

Dấu "=" xảy ra khi \(\begin{cases}a=1\\b=-3\\\end{cases}\)

\(A=5-8x+x^2=-8x+x^2+6-11\)

\(=\left(x-4\right)^2-11\)

Vì \(\left(x-4\right)^2\ge0\forall x\)\(\Rightarrow\left(x-4\right)^2-11\ge-11\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-4\right)^2=0\Leftrightarrow x-4=0\Leftrightarrow x=4\)

Vậy Amin = - 11 <=> x = 4

\(B=\left(2-x\right)\left(x+4\right)=-x^2-2x+8\)

\(=-\left(x^2+2x+1\right)+9=-\left(x+1\right)^2+9\)

Vì \(\left(x+1\right)^2\ge0\forall x\)\(\Rightarrow-\left(x+1\right)^2+9\le9\)

Dấu "=" xảy ra \(\Leftrightarrow-\left(x+1\right)^2=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

Vậy Bmax = 9 <=> x = - 1

a: Ta có: \(A=16x^2+8x+5\)

\(=16x^2+8x+1+4\)

\(=\left(4x+1\right)^2+4\ge4\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{1}{4}\)

b: Ta có: \(B=2x^2-5x\)

\(=2\left(x^2-\dfrac{5}{2}x\right)\)

\(=2\left(x^2-2\cdot x\cdot\dfrac{5}{4}+\dfrac{25}{16}-\dfrac{25}{16}\right)\)

\(=2\left(x-\dfrac{5}{4}\right)^2-\dfrac{25}{8}\ge-\dfrac{25}{8}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{5}{4}\)

Lời giải:

$A=(x^2+4y^2+4xy)+x^2+5-8x-12y$

$=(x+2y)^2-6(x+2y)+x^2+5-2x$

$=(x+2y)^2-6(x+2y)+9+(x^2-2x+1)-5$

$=(x+2y-3)^2+(x-1)^2-5\geq 0+0-5=-5$

Vậy $A_{\min}=-5$. Giá trị này đạt được khi $x+2y-3=x-1=0$

$\Leftrightarrow x=1; y=1$

\(A=x^2-8x+5\)

\(=\left(x^2-8x+16\right)-11\)

\(=\left(x-4\right)^2-11\)

\(=-11+\left(x-4\right)^2\)

Vì \(\left(x-4\right)^2\) ≥ 0

⇒ A ≥ -11

Min A=-11 ⇔\(x-4=0\)

                 ⇔\(x=4\)

toán 12 nha