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

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 \(x+2=0\Leftrightarrow x=-2\).

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 = - 4\(x\)² + 5\(x\) - 3

A = -( 4\(x^2\) - 5\(x\) + \(\dfrac{25}{16}\))  - \(\dfrac{23}{16}\)

A = -( 2\(x\) - \(\dfrac{5}{4}\))² - \(\dfrac{23}{16}\)

Vì ( 2\(x\) - \(\dfrac{5}{4}\))² ≥ 0; ⇒ - ( 2\(x\) - \(\dfrac{5}{4}\))² ≤ 0 ⇒ -( 2 \(x\) - \(\dfrac{5}{4}\))² - \(\dfrac{23}{16}\) ≤ - \(\dfrac{23}{16}\)

A(max) = - \(\dfrac{23}{16}\) ⇔ 2\(x\) - \(\dfrac{5}{4}\) = 0 ⇔ \(x\) = \(\dfrac{5}{4}\): 2 = \(\dfrac{5}{8}\)

Kết luận giá trị lớn nhất của biểu thức là - \(\dfrac{23}{16}\) xáy ra khi \(x\) = \(\dfrac{5}{8}\)

giá trị nhỏ nhất chứ hình như sai đề

\(giải:\)

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

\(=-4x^2+5x-\frac{25}{16}+\frac{41}{16}\)

\(=\left(-4x^2+5x-\frac{25}{16}\right)+\frac{41}{16}\)

\(=-\left(4x^2-5x+\frac{25}{16}\right)+\frac{41}{16}\)

\(=-\left[\left(2x\right)^2-2.2x.\frac{5}{4}+\left(\frac{5}{4}\right)^2\right]+\frac{41}{16}\)

\(=-\left(2x-\frac{5}{4}\right)^2+\frac{41}{16}\le\frac{41}{16}\)

\(GTLN\) \(của\)\(-4x^2+5x+1=\frac{41}{16}\)\(đạt\)\(khi\)\(-\left(2x-\frac{5}{4}\right)^2=0\)

                                                                                          \(\Leftrightarrow2x-\frac{5}{4}=0\)

                                                                                          \(\Leftrightarrow2x=\frac{5}{4}\Leftrightarrow x=\frac{5}{8}\)

vậy gtln của -4x^2+5x+1 bằng 41/16 tại x=5/8

Bài này giá trị lớn nhất nhé.

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

\(A=-\left(4x^2-5x-1\right)\)

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

\(A=-\left(\left(2x-\frac{5}{4}\right)^2-\frac{41}{16}\right)\)

\(A=-\left(2x-\frac{5}{4}\right)^2+\frac{41}{16}\le\frac{41}{16}\)

Vậy \(MaxA=\frac{41}{16}\Leftrightarrow-\left(2x-\frac{5}{4}\right)^2=0\)

                                  \(\Leftrightarrow x=\frac{5}{8}\)

1/

a, \(A=4x^2-4x+5=4x^2-4x+1+4=\left(2x-1\right)^2+4\ge4\)

Dấu "=" xảy ra khi x=1/2

Vậy Amin=4 khi x=1/2

b, \(B=3x^2+6x-1=3\left(x^2+2x+1\right)-4=3\left(x+1\right)^2-4\ge-4\)

Dấu "=" xảy ra khi x=-1

Vậy Bmin = -4 khi x=-1

2/

a, \(A=10+6x-x^2=-\left(x^2-6x+9\right)+19=-\left(x-3\right)^2+19\le19\)

Dấu "=" xảy ra khi x=3

Vậy Amax = 19 khi x=3

b, \(B=7-5x-2x^2=-2\left(x^2-\frac{5}{2}x+\frac{25}{16}\right)+\frac{31}{8}=-2\left(x-\frac{5}{4}\right)^2+\frac{31}{8}\le\frac{31}{8}\)

Dấu "=" xảy ra khi x=5/4

Vậy Bmax = 31/8 khi x=5/4

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

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

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

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

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

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

a:Ta có: \(A=-4x^2+x-1\)

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

\(=-4\left(x^2-2\cdot x\cdot\dfrac{1}{8}+\dfrac{1}{64}+\dfrac{63}{64}\right)\)

\(=-4\left(x-\dfrac{1}{8}\right)^2-\dfrac{63}{16}\le-\dfrac{63}{16}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{8}\)

b: Ta có: \(B=-3x^2+5x+6\)

\(=-3\left(x^2-\dfrac{5}{3}x-2\right)\)

\(=-3\left(x^2-2\cdot x\cdot\dfrac{5}{6}+\dfrac{25}{36}-\dfrac{97}{36}\right)\)

\(=-3\left(x-\dfrac{5}{6}\right)^2+\dfrac{97}{12}\le\dfrac{97}{12}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{5}{6}\)

c: Ta có: \(C=-x^2+3x+4\)

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

\(=-\left(x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{25}{4}\right)\)

\(=-\left(x-\dfrac{3}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)

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