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)

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

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

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

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

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

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

\(a,P=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}\\ P=-\dfrac{3}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\\ \Leftrightarrow3\sqrt{x}+6=10\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\left(tm\right)\)

Ta có : \(P=3A+2B\)

\(=\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{3}{\sqrt{x}+2}=\dfrac{2\sqrt{x}+3}{\sqrt{x}+2}.\)

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

Do \(x\ge0\Rightarrow\sqrt{x}+2\ge0\)

\(\Rightarrow-\dfrac{1}{\sqrt{x}+2}\ge-1\)

\(\Rightarrow P=2-\dfrac{1}{\sqrt{x}+2}\ge-1+2=1.\)

Vậy : \(MinP=1.\) Dấu đẳng thức xảy ra khi và chỉ khi \(x=0.\)

Ta có: \(P=A\cdot B\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

Lời giải:
\(P=\frac{2(\sqrt{x}+1)-3}{\sqrt{x}+1}=2-\frac{3}{\sqrt{x}+1}\)

Vì $\sqrt{x}\geq 0$ với mọi $x\neq 1; x\geq 0$

$\Rightarrow \sqrt{x}+1\geq 1\Rightarrow \frac{3}{\sqrt{x}+1}\leq 3$

$\Rightarrow P\geq 2-3=-1$
Vậy $P_{\min}=-1$. Giá trị này đạt tại $x=0$

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

Em kiểm tra lại đề, mẫu số của phân số đầu tiên chắc chắn bị sai