Không có max

`a)sqrt{x^2-2x+5}`

`=sqrt{x^2-2x+1+4}`

`=sqrt{(x-1)^2+4}`

Vì `(x-1)^2>=0`

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

`=>sqrt{(x-1)^2+4}>=sqrt4=2`

Dấu "=" xảy ra khi `x=1.`

`b)2+sqrt{x^2-4x+5}`

`=2+sqrt{x^2-4x+4+1}`

`=2+sqrt{(x-2)^2+1}`

Vì `(x-2)^2>=0`

`=>(x-2)^2+1>=1`

`=>sqrt{(x-2)^2+1}>=1`

`=>sqrt{(x-2)^2+1}+2>=3`

Dấu "=" xảy ra khi `x=2`

Câu 2 : Bạn cần thêm điều kiện a,b là các số không âm 

Áp dụng bất đẳng thức Cosi, ta có : \(12=3a+5b\ge2.\sqrt{3a.5b}=2\sqrt{15ab}\Rightarrow ab\le\frac{6^2}{15}=\frac{12}{5}\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}3a+5b=12\\3a=5b\\a,b\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}a=2\\b=\frac{6}{5}\end{cases}}\)

Vậy Max B = \(\frac{12}{5}\Leftrightarrow\hept{\begin{cases}a=2\\b=\frac{6}{5}\end{cases}}\)

\(D=\frac{4x+3}{x^2+1}\)

Min D : 

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

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

Ta thấy : \(\frac{\left(x+2\right)^2}{x^2+1}\ge0\forall x\)

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

Dấu "=" xảy ra khi \(x+2=0\Leftrightarrow x=-2\)

Max D : 

\(D=\frac{4x+3}{x^2+1}=\frac{-4x^2+4x-1+4x^2+4}{x^2+1}\)

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

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

Ta thấy : \(\frac{-\left(2x-1\right)^2}{x^2+1}\le0\forall x\)

\(\Rightarrow D=\frac{-\left(2x-1\right)^2}{x^2+1}+4\le4\)

Dấu "=" xảy ra khi \(2x-1=0\Leftrightarrow x=\frac{1}{2}\)

Lời giải:

$-F=5x^2+4x-3=5(x^2+\frac{4}{5}x+\frac{2^2}{5^2})-\frac{19}{5}$

$=5(x-\frac{2}{5})^2-\frac{19}{5}\geq \frac{-19}{5}$ với mọi $x$

$\Rightarrow F\leq \frac{19}{5}$

Vậy $F_{\max}=\frac{19}{5}$. Giá trị này đạt tại $x-\frac{2}{5}=0\Leftrightarrow x=\frac{2}{5}$

B = 2\(x^2\) - 4\(x\) - 8

B = 2(\(x^2\) - 2\(x\) + 4)  - 16

B = 2(\(x-2\))² - 16 

Vì (\(x-2\))² ≥ 0 ∀ \(x\) ⇒ 2(\(x-2\))² ≥ 0 ∀ \(x\)

⇒ 2(\(x-2\))²  - 16 ≥ -16 ∀ \(x\)

Dấu bằng xảy ra khi  (\(x-2\))² = 0 ⇒ \(x-2=0\) ⇒ \(x=2\)

Vậy B_min = -16 khi \(x=2\)

Tìm min của C biết:

C = \(x^2\) - 2\(xy\) + 2y² + 2\(x\) - 10y + 17

C = (\(x^2\) - 2\(xy\) + y²) + 2(\(x\) - y) + y² - 8y + 16 + 1

C = (\(x\) - y)² + 2(\(x\) - y) + 1  + (y² - 8y + 16) 

C = (\(x-y+1\))² + (y - 4)² 

Vì (\(x\) - y + 1)² ≥ 0 ∀ \(x;y\); (y - 4)² ≥ 0 ∀ y

Dấu bằng xảy ra khi: \(\left\{{}\begin{matrix}x-y+1=0\\y-4=0\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x-y+1=0\\y=4\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}x-4+1=0\\y=4\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=-1+4\\y=4\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)

Vậy C_min = 0 khi (\(x;y\)) = (3; 4)

D = \(x^2\) - \(xy\) + y² - 2\(x\) - 2y

D=[\(x^2\)-2\(x\)\(\dfrac{y}{2}\)+(\(\dfrac{y}{2}\))²]-(2\(x\)-2\(\dfrac{y}{2}\)) +1 +(\(\dfrac{3}{4}\)y²-2.\(\dfrac{\sqrt{3}}{2}\)y .\(\sqrt{3}\) +3) - 4

D = (\(x-\dfrac{y}{2}\))² - 2(\(x-\dfrac{y}{2}\))+ 1 + (\(\dfrac{\sqrt{3}}{2}\)y - \(\sqrt{3}\))² - 4

D = (\(x-\dfrac{y}{2}\) - 1)² + (\(\dfrac{\sqrt{3}}{2}\)y - \(\sqrt{3}\))² - 4

Vì (\(x-\dfrac{y}{2}\) - 1)² ≥  0 ∀ \(x\);y và (\(\dfrac{\sqrt{3}}{2}\)y - \(\sqrt{3}\))² ≥ 0 ∀ y 

Vậy (\(x-\dfrac{y}{2}\) - 1)² + (\(\dfrac{\sqrt{3}}{2}\)y - \(\sqrt{3}\))² - 4 ≥ - 4 ∀ \(x;y\)

Dấu bằng xảy ra khi: \(\left\{{}\begin{matrix}x-\dfrac{y}{2}-1=0\\\dfrac{\sqrt{3}}{2}y-\sqrt{3}=0\end{matrix}\right.\)

      ⇒ \(\left\{{}\begin{matrix}x-\dfrac{y}{2}-1=0\\\sqrt{3}.\left(\dfrac{1}{2}y-1\right)=0\end{matrix}\right.\)

  ⇒ \(\left\{{}\begin{matrix}x=1+\dfrac{1}{2}y\\\dfrac{1}{2}y=1\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=1+1\\y=1:\dfrac{1}{2}\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)

Vậy D_min = - 4 khi (\(x;y\)) =(2; 2)

Ta có: \(Q=-x^2-2x+2021\)

\(=-\left(x^2+2x+1-2022\right)\)

\(=-\left(x+1\right)^2+2022\le2022\forall x\)

Dấu '=' xảy ra khi x=-1

\(Q=-\left(x^2+2x+1\right)+2022\)

\(Q=-\left(x+1\right)^2+2022\le2022\)

\(Q_{max}=2022\) khi \(x=-1\)