\(5-3x^2+6x=-3x^2+6x+5=-3\left(x^2-2x-5\right)\)

\(=-3\left(x^2-2x+1-6\right)\)

\(=-3\left(x^2-2x+1\right)+18\)

\(=-3\left(x-1\right)^2+18\le18\forall x\)

Dấu = xảy ra khi: \(-3\left(x-1\right)^2=0\Leftrightarrow x=1\)

Vậy : GTLN là 18 tại x = 1

Nguyễn Hoàng Khánh Dương sai rồi nha bạn! Bạn thay x = 1 vào biểu thức xem có ra được giá trị MAX = 18 không???

Gọi biểu thức trên là A.Ta có: \(A=5-3x^2+6x=-3x^2+6x+5\)

\(=-3x^2+6x-3+8\)

\(=-3\left(x^2-2x+1\right)+8\)

\(=-3\left(x-1\right)^2+8\le8\) (do \(-3\left(x-1\right)^2\le0\forall x\))

Dấu "=" xảy ra \(\Leftrightarrow-3\left(x-1\right)^2=0\Leftrightarrow x=1\)

Vậy \(A_{max}=8\Leftrightarrow x=1\)

\(5-3x^2+6x\)

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

\(=-3\left[\left(x^2-2\cdot x\cdot1+1\right)-\frac{8}{3}\right]\)

\(=-3\left[\left(x-1\right)^2-\frac{8}{3}\right]\)

\(=-3\left(x-1\right)^2+8\)

\(=8-3\left(x-1\right)^2\le8\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x-1=0\Leftrightarrow x-1\)

Từ đây suy ra tth đúng :))

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

\(=x^2-2x+1-\left(4x^2+12x+9\right)+5\)

\(=-3x^2-14x-3\)

\(=-3\left(x^2+\frac{14}{3}x+\frac{49}{9}\right)+\frac{40}{3}\)

\(=-3\left(x+\frac{7}{3}\right)^2\le0\forall x\) 

Dau '' = '' xay ra \(\Leftrightarrow x=\frac{-7}{3}\)

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

\(=x^2-2x+1-\left(4x^2+12x+9\right)+5\)

\(=-3x^2-14x-3=-3\left(x^2+\frac{14}{3}x\right)-3\)

\(=-3\left(x^2+2.\frac{7}{3}x+\frac{49}{9}-\frac{49}{9}\right)-3\)

\(=-3\left(x+\frac{7}{3}\right)^2+\frac{40}{3}\le\frac{40}{3}\)

Dấu ''='' xảy ra khi x = -7/3 

Vậy GTLN của F bằng 40/3 tại x = -7/3 

1, Ta có: \(A=3x^2+8x+9=3\left(x^2+\frac{8}{3}x+3\right)=3\left(x^2+\frac{8}{3}x+\frac{16}{9}+\frac{11}{9}\right)\)

\(=3\left(x+\frac{4}{3}\right)^2+\frac{11}{3}\ge\frac{11}{3}\forall x\)

=> Min A = 11/3 tại x = -4/3

2, Ta có: \(A=-2x^2+6x+3=-2\left(x^2-3x-\frac{3}{2}\right)=-2\left(x^2-3x+\frac{9}{4}-\frac{15}{4}\right)\)

\(=-2\left(x-\frac{3}{2}\right)^2+\frac{15}{2}\le\frac{15}{2}\forall x\)

=> Max A = 15/2 tại x = 3/2

=.= hk tốt!!

Cảm ơn

\(A=x^2-6x+10\)

\(\Leftrightarrow A=x^2-2\cdot x\cdot3+3^2-9+10\)

\(\Leftrightarrow A=\left(x-3\right)^2+1\ge1\)     \(\forall x\in z\)

\(\Leftrightarrow A_{min}=1khix=3\)

\(B=3x^2-12x+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x\right)^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2-12+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x-2\sqrt{3}\right)^2-11\ge-11\)    \(\forall x\in z\)

\(\Leftrightarrow B_{min}=-11khix=2\)

\(6-2\left|1+3x\right|\le6\)'

Max \(A=6\Leftrightarrow1+3x=0\)

\(\Rightarrow3x=-1\)

\(\Rightarrow x=\frac{-1}{3}\)

\(\left|x-2\right|+\left|x-5\right|\ge0\)

Max \(B=0\Leftrightarrow\hept{\begin{cases}x-2=0\\x-5=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\x=5\end{cases}}}\)

A= 6-2|1+3x|

Amax khi và chỉ khi 2-/1+3x/min.Vì /1+3x/luôn lớn hơn hoạc bằng 0 mà 2/1-3x/min khi /1-3x/min.

=>để 2/1-3x/min thì /1-3x/=0 khi đó thì 2/1-3x/=0.A= 6-2|1+3x|=6-0=6

Vậy Amax= 6

\(E=\dfrac{\dfrac{5}{2}\left(2x^2+3\right)+\dfrac{15}{2}}{2x^2+3}=\dfrac{5}{2}+\dfrac{15}{2\left(2x^2+3\right)}\)

Do \(2x^2+3\ge3;\forall x\Rightarrow\dfrac{15}{2\left(2x^2+3\right)}\le\dfrac{15}{2.3}=\dfrac{5}{2}\)

\(\Rightarrow E\le\dfrac{5}{2}+\dfrac{5}{2}=5\)

\(E_{max}=5\) khi \(x=0\)

a) Đặt \(A=10+2x-5x^2\)

\(-A=5x^2-2x-10\)

\(-5A=25x^2-10x-50\)

\(-5A=\left(25x^2-10x+1\right)-51\)

\(-5A=\left(5x-1\right)^2-51\)

Do \(\left(5x-1\right)^2\ge0\forall x\)

\(\Rightarrow-5A\ge-51\)

\(A\le\frac{51}{5}\)

Dấu "=" xảy ra khi : \(5x-1=0\Leftrightarrow x=\frac{1}{5}\)

Vậy Max A = \(\frac{51}{5}\Leftrightarrow x=\frac{1}{5}\)

b) Đặt \(B=x^2-6x+10\)

\(B=\left(x^2-6x+9\right)+1\)

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

Mà \(\left(x-3\right)^2\ge0\forall x\)

\(B\ge1\)

Dấu "=" xảy ra khi :

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

Vậy Min B \(=1\Leftrightarrow x=3\)

D = \(-\dfrac{5}{x^2-4x+7}\)

Vì: x² - 4x + 7

= x² - 4x + 4 + 3

= (x - 2)² + 3 \(\ge\) 3 \(\forall\)x

\(\Rightarrow\) \(\dfrac{5}{\left(x-2\right)^2+3}\) \(\le\) \(\dfrac{5}{3}\) \(\forall\)x

\(\Rightarrow\) \(-\dfrac{5}{\left(x-2\right)^2+3}\)\(\ge\)-\(\dfrac{5}{3}\) \(\forall\)x

Dấu"=" xảy ra khi:

x - 2 = 0

\(\Rightarrow\) x = 2

Vậy.............

E = \(\dfrac{2x^2+4x+4}{x^2+2x+4}\)

Ta có:

\(\dfrac{2x^2+4x+4}{x^2+2x+4}\)

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

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

Vì:

x² + 2x + 4

= x² + 2x + 1 + 3

= (x + 1)² + 3 \(\ge\) 3 \(\forall\)x

\(\Rightarrow\) \(\dfrac{4}{\left(x+1\right)^2+3}\) \(\le\) \(\dfrac{4}{3}\) \(\forall\)x

\(\Rightarrow\) 2 - \(\dfrac{4}{\left(x+1\right)^2+3}\) \(\le\) \(\dfrac{2}{3}\) \(\forall\)x

Dấu "=" xảy ra khi:

x + 1 = 0

\(\Rightarrow\) x = -1

Vậy...............

F = \(\dfrac{6x+8}{x^2+1}\)

= \(\dfrac{x^2+6x+9-x^2-1}{x^2+1}\)

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

= \(\dfrac{\left(x+3\right)^2}{x^2+1}-1\) \(\ge\) -1 \(\forall\)x

Dấu "=" xảy ra khi:

(x + 3)² = 0

\(\Rightarrow\) x + 3 = 0

\(\Rightarrow\) x = -3

Vậy.....................