ta có :  (\(\sqrt{x}\)-   2   )\(^2\)\(\ge\)0

\(\Leftrightarrow\)x  -  4\(\sqrt{x}\)+  4  \(\ge\)0

\(\Leftrightarrow\)x  -  4\(\sqrt{x}\)+  4 +   8\(\sqrt{x}\) \(\ge\)8\(\sqrt{x}\)

   \(\Leftrightarrow\)(\(\sqrt{x}\)+    2  )\(^2\)\(\ge\)8\(\sqrt{x}\)

\(\Leftrightarrow\)-(\(\sqrt{x}\)+    2  )\(^2\)\(\le\)-8\(\sqrt{x}\)

\(\Leftrightarrow\)Q  \(\le\)\(\frac{-8\sqrt{x}}{\sqrt{x}}\)=   (   -  8  )

Dấu ''   =   ''   xaye ra tại   x =  4

\(\text{a)Để C đạt GTNN}\)

\(\Rightarrow\hept{\begin{cases}\left(x+2\right)^2\\\left(y-\frac{1}{5}\right)^2\end{cases}\ge0}\)

\(\Rightarrow\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2\ge0\)

\(\Rightarrow\left(x+2\right)^2+\left(y-\frac{1}{5}\right)^2-10\ge0-10\)

\(\Rightarrow C\ge-10\)

\(\text{Vậy minC=-10 khi x=-2;y= }\frac{1}{5}\)

b)\(\text{Để D đạt GTLN}\)

=>(2x-3)²+5 đạt GTNN

Mà (2x-3)²\(\ge\)5

\(\Rightarrow GTLN\)của \(A=\frac{4}{5}\)khi \(x=\frac{3}{2}\)

Ta có \(\frac{1}{P}=\frac{\left(x+yz\right)\left(y+zx\right)\left(z+xy\right)^2}{x^3y^3}=\frac{x+yz}{y}\cdot\frac{y+zx}{x}\cdot\frac{\left(z+xy\right)^2}{x^2y^2}\)

\(=\left(\frac{x}{y}+z\right)\left(\frac{y}{x}+z\right)\left(\frac{z}{xy}+1\right)^2=\left[1+\left(\frac{x}{y}+\frac{x}{y}\right)z+x^2\right]\left(\frac{z}{xy}+1\right)^2\ge\left(1+2x+x^2\right)\)\(\left[\frac{4x}{\left(x+y\right)^2}+1\right]^2\)\(=\left(z+1\right)^2\left[\frac{4z}{\left(z-1\right)^2}+1\right]^2=\left[\frac{4z\left(z+1\right)}{\left(z-1\right)^2}+1\right]^2=\left[6+\frac{12}{z-1}+\frac{8}{\left(z-1\right)^2}+z-1\right]^2\)

\(=\left[6+\frac{12}{z-1}+\frac{3\left(z-1\right)}{4}+\frac{8}{\left(z-1\right)^2}+\frac{z-1}{8}+\frac{z-1}{8}\right]\)

Áp dụng BĐT Cosi ta có:

\(\frac{1}{P}\ge\left[6+2\sqrt{\frac{12}{z-1}\cdot\frac{3\left(z-1\right)}{3}}+3\sqrt[3]{\frac{8}{\left(z-1\right)^2}\cdot\frac{z-1}{8}\cdot\frac{z-1}{8}}\right]^2=\frac{729}{4}\)

\(\Rightarrow P\le\frac{4}{729}\). dấu "=" xảy ra <=> \(\hept{\begin{cases}x=y=2\\z=5\end{cases}}\)

Để \(T_{max}=\frac{-2\left|x-2018\right|-2021}{2020+\left|x-2018\right|}\)

Thì \(2020+\left|x-2018\right|_{min}\)

và \(-2\left|x-2018\right|-2021_{max}\)

Mà \(\left|x-2018\right|\ge0\forall x\Rightarrow-2\left|x-2018\right|\le0\) 

\(\Rightarrow T_{max}\Leftrightarrow\left|x-2018\right|_{min}\)

\(\Rightarrow T_{max}=-\frac{2021}{2020}\Leftrightarrow\left|x-2018\right|=0\Leftrightarrow x=0\)

\(\)

RIM LM ĐÚNG NHƯNG SAI KQ NHÁ X = 2018

\(A=\frac{x}{\left(x+2018\right)^2}\Leftrightarrow\frac{1}{A}=\frac{\left(x+2018\right)^2}{x}\)\(=\frac{x^2+2.2018x+2018^2}{x}\)

\(=x+4036+\frac{2018^2}{x}\)

Vì \(x+\frac{2018^2}{x}\ge2\sqrt{x.\frac{2018^2}{x}=4036}\)

Vậy GTNN của \(\frac{1}{A}\)=4036+4036=8072

Vậy GTLN của A=\(\frac{1}{8072}\)

Với \(x< 0\Rightarrow A< 0\) (1)

Với \(x=0\Rightarrow A=0\) (2)

Với \(x>0\Rightarrow A>0\) (3)

Từ (1), (2), (3) ta thấy GTLN của A nếu có sẽ xảy ra tại các giá trị x dương

Xét \(x>0\) chia cả tử và mẫu của A cho x:

\(A=\frac{x}{x^2+2.2018x+2018^2}=\frac{1}{x+\frac{2018^2}{x}+2.2018}\)

\(\Rightarrow A\le\frac{1}{2\sqrt{x.\frac{2018^2}{x}}+2.2018}=\frac{1}{2.2018+2.1028}=\frac{1}{4.2018}=\frac{1}{8072}\)

\(\Rightarrow A_{max}=\frac{1}{8072}\) khi x=2018

bạn đặt ĐKXĐ và rút gọn P đi\(\sqrt{x}-x=-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4},\forall x\ne1\)

\(\Rightarrow Maxp=\frac{1}{4}\Leftrightarrow dấu=xảyra\)

\(\Leftrightarrow x=\frac{1}{4}\)