Lời giải:
\(A=\frac{x^2+x+1}{x^2+2x+1}=\frac{x^2+2x+1-x}{x^2+2x+1}=1-\frac{x}{x^2+2x+1}=1-\frac{x}{(x+1)^2}\)

Ta thấy \((x+1)^2-4x=x^2-2x+1=(x-1)^2\geq 0\)

\(\Rightarrow (x+1)^2\geq 4x\Rightarrow \frac{x}{(x+1)^2}\leq \frac{x}{4x}=\frac{1}{4}\)

\(\Rightarrow A=1-\frac{x}{(x+1)^2}\geq 1-\frac{1}{4}=\frac{3}{4}\)

Vậy \(A_{\min}=\frac{3}{4}\Leftrightarrow (x-1)^2=0\Leftrightarrow x=1\), tức là A đạt min khi $x=1$

Ta có: \(Q=\dfrac{x^2+x+1}{x^2+2x+1}\)

\(\Rightarrow\dfrac{1}{Q}=\dfrac{x^2+2x+1}{x^2+x+1}\)

Để Q min thì \(\dfrac{1}{Q}\) max

\(\dfrac{1}{Q}=\dfrac{x^2+2x+1}{x^2+x+1}=1+\dfrac{x}{x^2+x+1}\)

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

( Vì mẫu > 0 và tử \(\ge0\) )

\(\Rightarrow\dfrac{1}{Q}\) đạt GTNN là \(\dfrac{4}{3}\) khi x =1

Vậy Q đạt GTNN là \(\dfrac{3}{4}\) khi x = 1

Ta có: \(\dfrac{a+b}{a}=\dfrac{a}{b}\)

\(\Leftrightarrow\dfrac{a}{b}-1-\dfrac{1}{\dfrac{a}{b}}=0\)

\(\Leftrightarrow\left(\dfrac{a}{b}\right)^2-\dfrac{a}{b}-1=0\)

\(\Leftrightarrow\left(\dfrac{a}{b}-\dfrac{1}{2}\right)^2=\dfrac{5}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{a}{b}=\dfrac{\sqrt{5}+1}{2}\\\dfrac{a}{b}=\dfrac{-\sqrt{5}+1}{2}\end{matrix}\right.\)

Thế \(\dfrac{a}{b}\) vào PT \(x^2-x-1\)

\(\Rightarrowđpcm\)

$A=2x-\sqrt{x}=2(x-\frac{1}{2}\sqrt{x}+\frac{1}{4^2})-\frac{1}{8}$

$=2(\sqrt{x}-\frac{1}{4})^2-\frac{1}{8}$

$\geq \frac{-1}{8}$

Vậy $A_{\min}=-\frac{1}{8}$. Giá trị này đạt tại $x=\frac{1}{16}$

$B=x+\sqrt{x}$

Vì $x\geq 0$ nên $B\geq 0+\sqrt{0}=0$

Vậy $B_{\min}=0$. Giá trị này đạt tại $x=0$

Vì $2-x\geq 0$ (theo ĐKXĐ) nên $C=1+\sqrt{2-x}\geq 1$

Vậy $C_{\min}=1$. Giá trị này đạt tại $2-x=0\Leftrightarrow x=2$

\(ĐKXĐ:\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

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

\(P\left(x\right)=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)

\(P\left(x\right)=x-\sqrt{x}-2\sqrt{x}-2+2\sqrt{x}+2\)

\(P\left(x\right)=x-\sqrt{x}\)

Ta có : \(\dfrac{P\left(x\right)}{2020\sqrt{x}}=\dfrac{x-\sqrt{x}}{2020\sqrt{x}}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{2020\sqrt{x}}=\dfrac{\sqrt{x}-1}{2020}\)

Để \(\dfrac{P\left(x\right)}{2020\sqrt{x}}min\Leftrightarrow\dfrac{\sqrt{x}-1}{2020}min\Leftrightarrow\sqrt{x}-1\) min (vì 2020 > 0)

Lại có : \(\sqrt{x}-1\ge-1\forall x\)

Dấu "=" xảy ra <=> x = 0

Vậy Min\(\dfrac{P\left(x\right)}{2020\sqrt{x}}=\dfrac{-1}{2020}\Leftrightarrow x=0\)

\(P=\frac{1}{x^2+2x+6}\)

\(P=\frac{1}{\left(x+1\right)^2+5}\ge\frac{1}{5}\forall x\)

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

Vậy P_min = 1/5 khi và chỉ khi x = -1

ta có : \(x^2+2x+6=x^2+2x+1+5.\)

\(\Rightarrow\left(x+1\right)^2+5\)

ta có : \(\left(x+1\right)^2\ge0\)

\(\Rightarrow\left(x+1\right)^2+5\ge5\)

\(\Rightarrow\frac{1}{x^2+2x+6}\ge\frac{1}{5}\)

Vậy GTLN(P) = 1/5 khi x = -1 

Lời giải:

$x^2+2x+6=(x^2+2x+1)+5=(x+1)^2+5\geq 5$ với mọi $x\in\mathbb{R}$

Do đó: $P=\frac{1}{x^2+2x+6}\leq \frac{1}{5}$

Vậy $P_{\max}=\frac{1}{5}$. Giá trị đạt tại $x=-1$

\(P=\dfrac{1}{\left(x+1\right)^2+5}\le\dfrac{1}{5}\)

\(P_{max}\) khi \(x+1=0\Leftrightarrow x=-1\)

a) \(P=\dfrac{x^2-\sqrt[]{x}}{x+\sqrt[]{x}+1}-\dfrac{2x+\sqrt[]{x}}{\sqrt[]{x}}+\dfrac{2\left(x+\sqrt[]{x}-2\right)}{\sqrt[]{x}-1}\)

Điều kiện xác định \(\Leftrightarrow\left\{{}\begin{matrix}x>0\\\sqrt[]{x}-1\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

\(\Rightarrow P=\dfrac{\sqrt[]{x}\left[\left(\sqrt[]{x}\right)^3-1\right]}{x+\sqrt[]{x}+1}-\dfrac{\sqrt[]{x}\left(2\sqrt[]{x}+1\right)}{\sqrt[]{x}}+\dfrac{2\left(\sqrt[]{x}-1\right)\left(\sqrt[]{x}+2\right)}{\sqrt[]{x}-1}\)

\(\Rightarrow P=\dfrac{\sqrt[]{x}\left(\sqrt[]{x}-1\right)\left(x+\sqrt[]{x}+1\right)}{x+\sqrt[]{x}+1}-\left(2\sqrt[]{x}+1\right)+2\left(\sqrt[]{x}+2\right)\)

\(\Rightarrow P=\sqrt[]{x}\left(\sqrt[]{x}-1\right)-\left(2\sqrt[]{x}+1\right)+2\left(\sqrt[]{x}+2\right)\)

\(\Rightarrow P=x-\sqrt[]{x}-2\sqrt[]{x}-1+2\sqrt[]{x}+4\)

\(\Rightarrow P=x-\sqrt[]{x}+3\)

b) \(A=\dfrac{P}{2012\sqrt[]{x}}=\dfrac{x-\sqrt[]{x}+3}{2012\sqrt[]{x}}\)\(\)

\(=\dfrac{x-\sqrt[]{x}+\dfrac{1}{4}-\dfrac{1}{4}+3}{2012\sqrt[]{x}}\)

\(=\dfrac{\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2+\dfrac{11}{4}}{2012\sqrt[]{x}}\)

\(\Rightarrow A=\dfrac{\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2}{2012\sqrt[]{x}}+\dfrac{\dfrac{11}{4}}{2012\sqrt[]{x}}=\dfrac{\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2}{2012\sqrt[]{x}}+\dfrac{11}{4.2012\sqrt[]{x}}\)

Ta lại có  \(\dfrac{\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2}{2012\sqrt[]{x}}\ge0,\forall x\ne0\)

\(\dfrac{1}{\sqrt[]{x}}>0\Rightarrow\dfrac{11}{4.2012\sqrt[]{x}}\ge\dfrac{11}{4.2012}=\dfrac{11}{8048}\)

\(\Rightarrow A=\dfrac{\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2}{2012\sqrt[]{x}}+\dfrac{11}{4.2012\sqrt[]{x}}\ge\dfrac{11}{8048}\)

Dấu "=" xảy ra \(\Leftrightarrow\sqrt[]{x}=1\Leftrightarrow x=1\)

Vậy \(GTNN\left(A\right)=\dfrac{11}{8048}\left(tạix=1\right)\)

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

\(=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right).\left(\sqrt{x}+2\right)}{\sqrt{x}-1}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2.\left(\sqrt{x}+2\right)\)

\(=x-\sqrt{x}+3\)

b) \(\dfrac{P}{2012\sqrt{x}}=\dfrac{x-\sqrt{x}+3}{2012\sqrt{x}}=\dfrac{\sqrt{x}}{2012}-\dfrac{1}{2012}+\dfrac{3}{2012\sqrt{x}}\)

\(=\left(\dfrac{\sqrt{x}}{2012}+\dfrac{3}{2012\sqrt{x}}\right)-\dfrac{1}{2012}\)

\(\ge2\sqrt{\dfrac{\sqrt{x}.3}{2012^2\sqrt{x}}}-\dfrac{1}{2012}\) (BĐT Cauchy)

\(=\dfrac{2\sqrt{3}}{2012}-\dfrac{1}{2012}=\dfrac{2\sqrt{3}-1}{2012}\)

Dấu "=" xảy ra khi \(\dfrac{\sqrt{x}}{2012}=\dfrac{3}{2012\sqrt{x}}\Leftrightarrow x=3\)(tm)