-x^2+4x-5=-(x^2-4x+5)

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

=-(x-2)^2-1<=-1

=>6/-(x-2)^2-1>=-6

=>B<=6

Dấu = xảy ra khi x=2

c: \(=\left(x+1\right)^2+1>0\forall x\)

Trả lời:

a, \(x^2-6x+11=x^2-6x+9+2=\left(x-3\right)^2+2\ge2\forall x\)

Dấu "=" xảy ra khi x - 3 = 0 <=> x = 3

Vậy GTNN của biểu thức bằng 2 khi x = 3

b, \(-x^2+6x-11=-\left(x^2-6x+11\right)=-\left(x^2-6x+9+2\right)=-\left[\left(x-3\right)^2+2\right]\)

\(=-\left(x-3\right)^2-2\le-2\forall x\)

Dấu "=" xảy ra khi x - 3 = 0 <=> x = 3

Vậy GTLN của biểu thức bằng - 2 khi x = 3

c, \(x^2+2x+2=x^2+2x+1+1=\left(x+1\right)^2+1\ge1>0\forall x\inℤ\)  (đpcm)

Dấu "=" xảy ra khi x + 1 = 0 <=> x = - 1

Mình nghĩ ra câu C rồi bạn nào giúp mình nghĩ nốt câu A,B hộ mình nhé mình cảm ơn!

a:6x-5-9x^2

=-(9x^2-6x+5)

=-(9x^2-6x+1+4)

=-(3x-1)^2-4<=-4

=>A>=2/-4=-1/2

Dấu = xảy ra khi x=1/3

b: \(B=\dfrac{4x^2-6x+4-1}{2x^2-3x+2}=2-\dfrac{1}{2x^2-3x+2}\)

2x^2-3x+2=2(x^2-3/2x+1)

=2(x^2-2*x*3/4+9/16+7/16)

=2(x-3/4)^2+7/8>=7/8

=>-1/2x^2-3x+2<=-1:7/8=-8/7

=>B<=-8/7+2=6/7

Dâu = xảy ra khi x=3/4

\(4x^2+4x+6\)

\(=\left(2x\right)^2+2.2x.1+1+5\)

\(=\left(2x+1\right)^2+5\ge5\)

\(Min=5\Leftrightarrow2x+1=0\Rightarrow x=\frac{-1}{2}\)

\(x^2+6x+11\)

\(=x^2+2.x.3+9+2\)

\(=\left(x+3\right)^2+2\ge2\)

\(Min=2\Leftrightarrow x+3=0\Rightarrow x-3\)

\(x^2-3x+1\)

\(=x^2-2.x.\frac{3}{2}+\frac{9}{4}-\frac{5}{4}\)

\(=\left(x+\frac{3}{2}\right)^2-\frac{5}{4}\le\frac{-5}{4}\)

\(MIn=\frac{-5}{4}\Leftrightarrow x+\frac{3}{2}=0\Rightarrow x=\frac{-3}{2}\)

B = 4x² + 4x - 6 = (2x)² + 2.2.x + 1 - 7 = (2x + 1)² - 7 \(\ge\)-7

             Vậy MinB = -7 khi 2x + 1 = 0 => x = -1/2 

C = x² + 6x + 11 = x² + 2.3.x + 9 + 2 = (x + 3)² + 2 \(\ge\)2

              Vậy MinC = 2 khi x + 3 = 0 => x = -3

D = x² - 3x + 1 \(=x^2-2.\frac{3}{2}.x+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2+1=\left(x-\frac{3}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\)

              Vậy MinD = -5/4 khi x - 3/2 = 0 => x = 3/2

bài a của  o0o I am a studious person o0o có lẽ sai

\(B=4x^2+4x-6=\left(4x^2+4x+1\right)-7=\left(2x+1\right)^2-7\)

có:\(\left(2x+1\right)^2\ge0\)

vậy GTNN của B = -7 tại x = -1/2

MinC = 3/4 (khi x = -3/2)

\(C=\dfrac{\left(x^2+3\right)\left(x^4-3x^2+9\right)}{x^4+3x^2-3x^3-9x+3x^2+9}=\dfrac{\left(x^2+3\right)\left(x^4+6x^2+9-9x^2\right)}{\left(x^2+3\right)\left(x^2-3x+3\right)}\\ C=\dfrac{\left(x^2+3\right)^2-9x^2}{x^2-3x+3}=\dfrac{\left(x^2-3x+3\right)\left(x^2+3x+3\right)}{x^2-3x+3}\\ C=x^2+3x+3=\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{3}{4}=\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Dấu \("="\Leftrightarrow x=-\dfrac{3}{2}\)

\(Q=-2\left(x-\dfrac{3}{2}\right)^2+\dfrac{25}{2}\le\dfrac{25}{2}\)

\(Q_{max}=\dfrac{25}{2}\) khi \(x=\dfrac{3}{2}\)

\(A=\dfrac{9\left(x^2+2\right)-9x^2+6x-1}{x^2+2}=9-\dfrac{\left(3x-1\right)^2}{x^2+2}\le9\)

\(A_{max}=9\) khi \(x=\dfrac{1}{3}\)

\(A=\dfrac{12x+34}{2\left(x^2+2\right)}=\dfrac{-\left(x^2+2\right)+x^2+12x+36}{2\left(x^2+2\right)}=-\dfrac{1}{2}+\dfrac{\left(x+6\right)^2}{2\left(x^2+2\right)}\le-\dfrac{1}{2}\)

\(A_{min}=-\dfrac{1}{2}\) khi \(x=-6\)

khai triển hằng đẳng thức số một và 2 bạn ơi 

a)\(x^2-6x+11\)

\(=\left(x^2-6x+9\right)+2=\left(x-3\right)^2+2\ge2\)

Dấu "="xảy ra khi x=3

b)\(-x^2+6x-11\)

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

\(=-\left(x-3\right)^2-2\le-2\)

Dấu "=" xảy ra khi x=3

a) \(A=6x-x^2-11=-\left(x^2-6x+9\right)-2=-\left(x-3\right)^2-2\le-2\)

Dấu \(=\)khi \(x-3=0\Leftrightarrow x=3\).

b) \(B=x^2-5x-2=x^2-2.\frac{5}{2}x+\left(\frac{5}{2}\right)^2-\frac{33}{4}=\left(x-\frac{5}{2}\right)^2-\frac{33}{4}\ge-\frac{33}{44}\)

Dấu \(=\)khi \(x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\).

\(A=x^2-6x+11\\ =x^2-2.3.x+3^2+2\\ =\left(x-3\right)^2+2\)

\(\left(x-3\right)^2\ge0\forall x\\ \Rightarrow\left(x-3\right)^2+2\ge2\forall x\)

Dấu = xảy ra khi

\(x-3=0\\ \Rightarrow x=3\)

Vậy \(Min_A=2\) khi \(x=3\)

\(A=x^2-6x+11\)
\(=\left(x^2-6x+9\right)+2\)
\(=\left(x-3\right)^2+2\ge2\)
Dấu"=" xảy ra khi \(x=3\)

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

\(=x^2-3x-3x+9+2\)

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

Mà \(\left(x-3\right)^2+2\ge0\)

Vậy GTLN của A Là 2 khi x=3