\(A=\frac{5x^2+4x-1}{x^2}=\frac{9x^2-\left(4x^2-4x+1\right)}{x^2}=9-\frac{\left(2x-1\right)^2}{x^2}\le9\)

Dấu \(=\)khi \(2x-1=0\Leftrightarrow x=\frac{1}{2}\).

\(B=\frac{x^2}{x^2+x+1}=\frac{3x^2}{3x^2+3x+3}=\frac{4x^2+4x+4-\left(x^2+4x+4\right)}{3x^2+3x+3}=\frac{4}{3}-\frac{\left(x+2\right)^2}{3\left(x^2+x+1\right)}\le\frac{4}{3}\)

Dấu \(=\)khi \(x+2=0\Leftrightarrow x=-2\).

A = x² - 4x + 1 

A = ( x² - 4x + 4 ) - 3

A = ( x - 2 )² - 3

( x - 2 )² ≥ 0 ∀ x => ( x - 2 )² - 3 ≥ -3

Đẳng thức xảy ra <=> x - 2 = 0 => x = 2

=> MinA = -3 <=> x = 2

B = 4x² + 4x + 11

B = 4( x² + x + 1/4 ) + 10

B = 4( x + 1/2 )² + 10

4( x + 1/2 )² ≥ 0 ∀ x => 4( x + 1/2 )² + 10 ≥ 10

Đẳng thức xảy ra <=> x + 1/2 = 0 => x = -1/2

=> MinB = 10 <=> x = -1/2

C = ( x - 1 )( x + 3 )( x + 2 )( x + 6 )

C = [ ( x - 1 )( x + 6 ) ][ ( x + 3 )( x + 2 ) ]

C = [ x² + 5x - 6 ][ x² + 5x + 6 ]

C = ( x² + 5x )² - 6² = ( x² + 5x )² - 36

( x² + 5x )² ≥ 0 ∀ x => ( x² + 5x )² - 36 ≥ -36

Đẳng thức xảy ra <=> x² + 5x = 0

                             <=> x( x + 5 ) = 0

                             <=> \(\orbr{\begin{cases}x=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)

=> MinC = -36 <=> x = 0 hoặc x = -5

D = 5 - 8x - x²

D = -( x² + 8x + 16 ) + 21

D = -( x + 4 )² + 21

-( x + 4 )² ≤ 0 ∀ x => -( x + 4 )² + 21 ≤ 21

Đẳng thức xảy ra <=> x + 4 = 0 => x = -4

=> MaxD = 21 <=> x = -4

E = 4x - x² + 1

E = -( x² - 4x + 4 ) + 5

E = -( x - 2 )² + 5

-( x - 2 )² ≤ 0 ∀ x => -( x - 2 )² + 5 ≤ 5 

Đẳng thức xảy ra <=> x - 2 = 0 => x = 2

=> MaxE = 5 <=> x = 2

a, \(A=x^2-6x+11\)

\(=x^2-2.3.x+9+2\)

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

Ta có: \(\left(x-3\right)^2\ge0\Leftrightarrow\left(x-3\right)^2+2\ge2\)

Dấu "=" xảy ra \(\Leftrightarrow x-3=0\)\(\Leftrightarrow x=3\)

Vậy \(MinA=3\Leftrightarrow x=3\)

b, \(B=2x^2+10x-1\)

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

\(=2\left(x^2+2.\frac{5}{2}x+\frac{25}{4}\right)-\frac{21}{4}\)

\(=2\left(x+\frac{5}{2}\right)^2-\frac{21}{4}\)

Ta có: \(\left(x+\frac{5}{2}\right)^2\ge0\Leftrightarrow\left(x+\frac{5}{2}\right)^2-\frac{21}{4}\ge-\frac{21}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x+\frac{5}{2}\right)^2=0\Leftrightarrow x+\frac{5}{2}=0\Leftrightarrow x=-\frac{5}{2}\)

Vậy \(MinB=-\frac{21}{4}\Leftrightarrow x=-\frac{5}{2}\)

c, \(C=5x-x^2\)

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

\(=-\left(x^2+2.\frac{5}{2}x+\frac{25}{4}\right)+\frac{25}{4}\)

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

Ta có: \(-\left(x+\frac{5}{2}\right)^2\le0\Leftrightarrow-\left(x+\frac{5}{2}\right)^2+\frac{25}{4}\le\frac{25}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x+\frac{5}{2}\right)^2=0\Leftrightarrow x=-\frac{5}{2}\)

Vậy \(MaxB=\frac{25}{4}\Leftrightarrow x=-\frac{5}{2}\)

Ta có: A = 2x² - 5x - 8 = 2(x² - 5/2x + 25/16) - 89/8 = 2(x - 5/4)² - 89/8

Ta luôn có: 2(x - 5/4)² \(\ge\)0 \(\forall\)x

=> 2(x - 5/4)² - 89/8 \(\ge\)-89/8 \(\forall\)x

Dấu "=" xảy ra <=> x - 5/4 = 0 <=> x = 5/4

Vậy Min của A = -89/8 tại x = 5/4

Ta có: B = -x² - 4x + 3 = -(x² + 4x + 4) + 7 = -(x + 2)² + 7

Ta luôn có: -(x + 2)² \(\le\)0 \(\forall\)x

=> -(x + 2)² + 7 \(\le\)7 \(\forall\)x

Dấu "=" xảy ra <=> x + 2 = 0 <=> x = -2

Vậy Max của B = 7 tại x = -2

Bài 2: 

a: Ta có: \(x^2+4x+7\)

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

\(=\left(x+2\right)^2+3\ge3\forall x\)

Dấu '=' xảy ra khi x=-2

Ta có:

\(B=-5x^2-4x+1\)

\(=\left(4x^2-4x+1\right)-9x^2\)

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

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

\(=-\left(x+1\right)\left(5x-1\right)\)

\(B=-5x^2-4x+1\)

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

\(B=-5\left[x^2+2.x.\frac{2}{5}+\left(\frac{2}{5}\right)^2-\frac{9}{25}\right]\)

\(B=-5\left(x+\frac{2}{5}\right)^2+5.\frac{9}{25}\)

\(B=-5\left(x+\frac{2}{5}\right)^2+\frac{9}{5}\)

Ta có: \(\left(x+\frac{2}{5}\right)^2\ge0\forall x\)

\(\Rightarrow-5.\left(x+\frac{2}{5}\right)^2\le0\forall x\)

\(\Rightarrow-5.\left(x+\frac{2}{5}\right)^2+\frac{9}{5}\le\frac{9}{5}\forall x\)

\(B=\frac{9}{5}\Leftrightarrow-5.\left(x+\frac{2}{5}\right)^2=0\Leftrightarrow x+\frac{2}{5}=0\Leftrightarrow x=-\frac{2}{5}\)

Vậy \(B_{max}=\frac{9}{5}\Leftrightarrow x=-\frac{2}{5}\)

Tham khảo nhé~

\(B=-5x^2-4x+1\)

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

\(=\left(-5x^2-4x-\frac{4}{5}\right)+\frac{9}{5}\)

\(=-5\left(x^2+\frac{4}{5}x+\frac{4}{25}\right)+\frac{9}{5}\)

\(=-5\left(x+\frac{2}{5}\right)^2+\frac{9}{5}\le\frac{9}{5}\)(vì \(-5\left(x+\frac{2}{5}\right)^2\le0\forall x\))

Dấu "=" xảy ra \(\Leftrightarrow-5\left(x+\frac{2}{5}\right)^2=0\Leftrightarrow x=-\frac{2}{5}\)

Vậy \(B_{max}=\frac{9}{5}\Leftrightarrow x=-\frac{2}{5}\)

\(A=-x^2+3x-5\)\(=-\dfrac{11}{4}-\left(x^2-2.\dfrac{3}{2}x+\dfrac{9}{4}\right)=-\dfrac{11}{4}-\left(x-\dfrac{3}{2}\right)^2\le-\dfrac{11}{4}\) với mọi x

\(\Rightarrow A_{max}=-\dfrac{11}{4}\Leftrightarrow x-\dfrac{3}{2}=0\Leftrightarrow x=\dfrac{3}{2}\)

\(B=5x-4x^2-3=-\dfrac{23}{16}-\left(4x^2-2.\dfrac{5}{4}.2x+\dfrac{25}{16}\right)\)\(=-\dfrac{23}{16}-\left(2x-\dfrac{5}{4}\right)^2\)\(\le-\dfrac{23}{16}\forall x\)

\(\Rightarrow B_{max}=-\dfrac{23}{16}\Leftrightarrow2x-\dfrac{5}{4}=0\Leftrightarrow x=\dfrac{5}{8}\)

\(C=5-4x-25x^2=\dfrac{129}{25}-\left(25x^2+2.5x.\dfrac{2}{5}+\dfrac{4}{25}\right)\)\(=\dfrac{129}{25}-\left(5x+\dfrac{2}{5}\right)^2\le\dfrac{129}{25}\forall x\)

\(\Rightarrow C_{max}=\dfrac{129}{25}\Leftrightarrow5x+\dfrac{2}{5}=0\Leftrightarrow x=-\dfrac{2}{25}\)

\(D=3x-2x^2=-2\left(x^2-\dfrac{3}{2}x\right)=-2\left(x^2-2.\dfrac{3}{4}x+\dfrac{9}{16}\right)+\dfrac{9}{8}\)\(=\dfrac{9}{8}-2\left(x-\dfrac{3}{4}\right)^2\le\dfrac{9}{8}\) với mọi x

\(\Rightarrow D_{max}=\dfrac{9}{8}\Leftrightarrow x-\dfrac{3}{4}=0\Leftrightarrow x=\dfrac{3}{4}\)

\(E=2+6x-\dfrac{1}{4}x^2=-\dfrac{1}{4}\left(x^2-24x\right)+2=-\dfrac{1}{4}\left(x^2-2.12x+144\right)+38\)\(=38-\dfrac{1}{4}\left(x-12\right)^2\le38\forall x\)

\(\Rightarrow E_{max}=38\Leftrightarrow x-12=0\Leftrightarrow x=12\)

\(F=-5x^2+4x=-5\left(x^2-\dfrac{4}{5}x\right)=-5\left(x^2-2.\dfrac{2}{5}x+\dfrac{4}{25}\right)+\dfrac{4}{5}\)\(=\dfrac{4}{5}-5\left(x-\dfrac{2}{5}\right)^2\le\dfrac{4}{5}\forall x\)

\(\Rightarrow F_{max}=\dfrac{4}{5}\Leftrightarrow x-\dfrac{2}{5}=0\Leftrightarrow x=\dfrac{2}{5}\)

Ta có: A = 4 - 5x² - y² + 2xy - 4x

A = -(5x² + y² .- 2xy + 4x - 4)

A = -[(x² - 2xy + y²) + (4x² + 4x + 1) - 5]

A = -(x - y)² - (2x + 1)² + 5 \(\ge\)5 \(\forall\)x;y

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x-y=0\\2x+1=0\end{cases}}\) => \(x=y=-\frac{1}{2}\)

Vậy MaxA = 5 <=> x = y = -1/2