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`

giúp mk vs 

lên mạng mà tra là ra mà

\(1,Sửa:A=4x^4+4x^2y+y^2+2=\left(2x^2+y\right)^2+2\ge2\\ A_{min}=2\Leftrightarrow2x^2+y=0\Leftrightarrow x^2=-\dfrac{y}{2}\\ 2,B=\left(x+y\right)^2+\left(y+1\right)^2+12\ge12\\ B_{min}=12\Leftrightarrow\left\{{}\begin{matrix}x=-y=1\\y=-1\end{matrix}\right.\)

\(\dfrac{M}{3}=\dfrac{x^2+y^2-xy}{x^2+y^2+xy}=\dfrac{3\left(x^2+y^2+xy\right)-2\left(x^2+y^2+2xy\right)}{x^2+y^2+xy}=3-\dfrac{2\left(x+y\right)^2}{x^2+y^2+xy}\le3\)

\(\Rightarrow M\le9\)

\(M_{max}=9\) khi \(\left\{{}\begin{matrix}x+y=0\\x^2+y^2+xy=3\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(-\sqrt{3};\sqrt{3}\right);\left(\sqrt{3};-\sqrt{3}\right)\)

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

\(\Rightarrow M\ge1\)

\(M_{min}=1\) khi \(\left\{{}\begin{matrix}x-y=0\\x^2+y^2+xy=3\end{matrix}\right.\) \(\Rightarrow x=y=\pm1\)

\(\dfrac{x^2+y^2}{2}\ge xy\Rightarrow-xy\ge-\dfrac{x^2+y^2}{2}\)

\(\Rightarrow4=x^2+y^2-xy\ge x^2+y^2-\dfrac{x^2+y^2}{2}=\dfrac{x^2+y^2}{2}\)

\(\Rightarrow x^2+y^2\le8\)

\(C_{max}=8\) khi \(x=y=\pm2\)

\(x^2+y^2\ge-2xy\Rightarrow-xy\le\dfrac{x^2+y^2}{2}\)

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

\(\Rightarrow x^2+y^2\ge\dfrac{8}{3}\)

\(C_{min}=\dfrac{8}{3}\) khi \(\left(x;y\right)=\left(-\dfrac{2}{\sqrt{3}};\dfrac{2}{\sqrt{3}}\right);\left(\dfrac{2}{\sqrt{3}};-\dfrac{2}{\sqrt{3}}\right)\)

Đúng thì like giúp mik nha bạn. Thx bạn

Chọn B

Chọn B