\(T=\sqrt{\dfrac{3\sqrt{x}}{\sqrt{x}-6}\cdot\dfrac{x-6\sqrt{x}}{\sqrt{x}-1}}=\sqrt{\dfrac{3\sqrt{x}}{\sqrt{x}-6}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-6\right)}{\sqrt{x}-1}}\\ =\sqrt{\dfrac{3\sqrt{x}\cdot\sqrt{x}}{\sqrt{x}-1}}=\sqrt{\dfrac{3x}{\sqrt{x}-1}}\\ =\sqrt{\dfrac{3\left(x-1\right)+3}{\sqrt{x}-1}}=\sqrt{3\left(\sqrt{x}+1\right)+\dfrac{3}{\sqrt{x}-1}}\\ =\sqrt{3\left(\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}\right)+6}\)

Áp dụng bất đẳng thức Cosi ta có:

\(\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}\ge2\)

\(\Rightarrow T\ge\sqrt{3\cdot2+6}=2\sqrt{3}\)

Dấu = xảy ra khi x=4

ĐKXĐ: \(\left\{{}\begin{matrix}2x-1\ge0\\x+2>0\\3-x\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x>-2\\x\le3\end{matrix}\right.\) 

\(\Rightarrow\dfrac{1}{2}\le x\le3\)

ĐKXĐ: 2x-1>=0 và \(x-\sqrt{2x-1}>0\)

=>x>=1/2 và x>căn 2x-1

=>x>=1/2 và x^2>2x-1

=>x>=1/2 và x^2-2x+1>0

=>x>=1/2 và x<>1

\(\dfrac{1}{\sqrt[]{x-\sqrt[]{2x-1}}}\left(1\right)\)

\(\left(1\right)xđ\Leftrightarrow\left\{{}\begin{matrix}x-\sqrt[]{2x-1}>0\\2x-1\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt[]{2x-1}< x\left(2\right)\\x\ge\dfrac{1}{2}\end{matrix}\right.\) \(\left(I\right)\)

\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\2x-1\ge0\\2x-1< x^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ge\dfrac{1}{2}\\x^2+2x-1>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ge\dfrac{1}{2}\\\left(x-1\right)^2>0,\forall x\ne0\end{matrix}\right.\)  \(\Leftrightarrow x\ge\dfrac{1}{2}\)

\(\left(I\right)\Leftrightarrow x\ge\dfrac{1}{2}\)

Đính chính

\(...\left(x-1\right)^2>0,\forall x\ne1\)

\(...\Rightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x\ne1\end{matrix}\right.\)

x>=1

2TH (mik dự đoán)

TH1: Nếu là \(\sqrt{x}-1\) => ĐKXĐ: x \(\ge\) 0

TH2: Nếu là \(\sqrt{x-1}\) => ĐKXĐ: \(x\ge1\)

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

ĐKXĐ: \(\left[{}\begin{matrix}x\ge\sqrt{5}\\x\le-\sqrt{5}\end{matrix}\right.\)

b: ĐKXĐ: \(\left[{}\begin{matrix}x\ge1\\x< -12\end{matrix}\right.\)

x∈[0, ∞)

\(a,ĐK:x\ne4;x\ge3\\ b,ĐK:x\ge1\)

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

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

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

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

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

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

b) Để P>0 thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}>0\)

mà \(\sqrt{x}+1>0\forall x\) thỏa mãn ĐKXĐ

nên \(\sqrt{x}\left(\sqrt{x}-1\right)>0\)

mà \(\sqrt{x}>0\forall x\) thỏa mãn ĐKXĐ

nên \(\sqrt{x}-1>0\)

\(\Leftrightarrow\sqrt{x}>1\)

hay x>1

Kết hợp ĐKXĐ,ta được: x>1

Vậy: Để P>0 thì x>1