a) \(A=4x^2-4x-1\)

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

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

\(\Rightarrow Min_A=-2\)

\(\Leftrightarrow x=\frac{1}{2}\)

Vậy ...

b) \(B=\frac{1}{4}x^2+x-1\)

\(=\left(\frac{1}{2}x\right)^2+2.\left(\frac{1}{2}x\right)+1-1-1\)

\(=\left(\frac{1}{2}x+1\right)^2-2\)

\(\Rightarrow Min_B=-2\)

\(\Leftrightarrow x=-2\)

Vậy ...

a) \(A=4x^2-4x-1\)

\(A=4x^2-4x+1-2\)

\(A=\left(2x-1\right)^2-2\) 

Có: \(\left(2x-1\right)^2\ge0\Rightarrow\left(2x-1\right)^2-2\ge-2\)

Dấu '=' xảy ra khi: \(\left(2x-1\right)^2=0\Rightarrow2x-1=0\Rightarrow x=\frac{1}{2}\)

Vậy: \(Min_A=-2\) tại \(x=\frac{1}{2}\)

b) \(B=\frac{1}{4}x^2+x-1\)

\(B=\frac{1}{4}x^2+x+1-2\)

\(B=\left(\frac{1}{2}x+1\right)^2-2\)

Có: \(\left(\frac{1}{2}x+1\right)^2\ge0\Rightarrow\left(\frac{1}{2}x+1\right)^2-2\ge-2\)

Dấu = xảy ra khi: \(\left(\frac{1}{2}x+1\right)^2=0\Rightarrow\frac{1}{2}x+1=0\Rightarrow x=-\frac{1}{2}\)

Vậy: \(Min_B=-2\) tại \(x=-\frac{1}{2}\)

bài b. x=-2

M=(8x+3)/(4x^2+1) 
M = ( - 4x^2 - 1 + 4x^2 + 8x + 4)/(4x^2 +1) 
M= -1 + (2x +2)^2/(4x^2 +1) ≥ -1 
=> min M = -1 khi x = -1 
mặt khác: 
M = -1 + (2x +2)^2/(4x^2 +1) 
M = 4 - 5 + (2x +2)^2/(4x^2 +1) 
M = 4 - ( 20x^2 + 5 - 4x^2 - 8x - 4)/(4x^2 +1) 
M = 4 - (16x^2 - 8x +1)/(4x^2 +1) 
M = 4 - (4x - 1)^2/(4x^2 +1) ≤ 4 
=> max M = 4 khi x = 1/4 

\(A=\frac{4x^2+8x+4-\left(4x^2+1\right)}{4x^2+1}=\frac{\left(2x+2\right)^2}{4x^2+1}-1\ge-1\)

\(A_{min}=-1\) khi \(x=-1\)

\(A=\frac{16x^2+4-\left(16x^2-8x+1\right)}{4x^2+1}=4-\frac{\left(4x-1\right)^2}{4x^2+1}\le4\)

\(A_{max}=4\) khi \(x=\frac{1}{4}\)

\(A=\frac{4x^2-12x+15}{x^2-3x+3}=4+\frac{3}{x^2-3x+3}=4+\frac{3}{\left(x-\frac{3}{2}\right)^2+\frac{3}{4}}\le8\)

dau '=' xay ra khi \(x=\frac{3}{2}\)

\(B=\frac{4x^2-8x+12}{x^2-2x+5}=4-\frac{8}{x^2-2x+5}=4-\frac{8}{\left(x-1\right)^2+4}\le2\)

dau '=' xay ra khi \(x=1\)

Tìm GTLN:

\(A=-x^2+6x-15\)

\(=-\left(x^2-6x+15\right)\)

\(=-\left(x^2-2.x.3+9+6\right)\)

\(=-\left(x+3\right)^2-6\le0\forall x\)

Dấu = xảy ra khi: 

   \(x-3=0\Leftrightarrow x=3\)

Vậy A_max = - 6 tại x = 3

Tìm GTNN :

\(A=x^2-4x+7\)

\(=x^2+2.x.2+4+3\)

\(=\left(x+2\right)^2+3\ge0\forall x\)

Dấu = xảy ra khi:

   \(x+2=0\Leftrightarrow x=-2\)

Vậy A_min = 3 tại x = - 2

Các câu còn lại làm tương tự nhé... :)

giải hết i

a: Ta có: \(A=2x^2-8x+1\)

\(=2\left(x^2-4x+\dfrac{1}{2}\right)\)

\(=2\left(x^2-4x+4-\dfrac{7}{2}\right)\)

\(=2\left(x-2\right)^2-7\ge-7\forall x\)

Dấu '=' xảy ra khi x=2

bạn làm rõ ra dc ko mik ko hiểu

\(4B=4x^2+4xy+4y^2-8x-12y+8076\)

= \(\left(2y\right)^2-4y\left(3-x\right)+\left(3-x\right)^2-\left(3-x\right)^2\)

\(+\left(2x\right)^2-8x+8076\)

= \(\left(2y-3+x\right)^2+3x^2-2x+8076\)

đến đây thì dễ rồi

1.(√x -2)^2 ≥ 0 --> x -4√x +4 ≥ 0 --> x+16 ≥ 12 +4√x --> (x+16)/(3+√x) ≥4 
--> Pmin=4 khi x=4

2. Đặt \(\sqrt{x^2-4x+5}=t\ge1\)1

=> M=2x²-8x+\(\sqrt{x^2-4x+5}\)+6=2(t²-5)+t+6

<=> M=2t²+t-4\(\ge\)2.1²+1-4=-1

M_min=-1 khi t=1 hay x=2