\(\sqrt{\dfrac{4-2x}{x^2}}\) có nghĩa thì \(\left\{{}\begin{matrix}4-2x\ge0\\x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x\le4\\x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le2\\x\ne0\end{matrix}\right.\)

ĐKXĐ: \(\dfrac{x^2}{x+1}>=0\)

=>x+1>0

=>x>-1

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

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

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

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

b: Thay \(x=3+2\sqrt{2}\) vào P, ta được:

\(P=\dfrac{2\sqrt{2}+2}{\sqrt{2}+1}=2\)

\(ĐKXĐ:\left\{{}\begin{matrix}\dfrac{x^2}{x+1}\ge0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\x\ne-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\ne-1\end{matrix}\right.\Leftrightarrow x>-1\)

\(\sqrt{\dfrac{4}{2x+3}}\) xác định khi \(\dfrac{4}{2x+3}\ge0\Rightarrow2x+3>0\Rightarrow x>-\dfrac{3}{2}\)

\(\sqrt{\dfrac{2x-1}{2-x}}\) xác định khi \(\dfrac{2x-1}{2-x}\ge0\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-1\ge0\\2-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-1\le0\\2-x< 0\end{matrix}\right.\left(l\right)\end{matrix}\right.\Rightarrow\dfrac{1}{2}\le x< 2\)

ĐKXĐ: \(-\dfrac{2}{x-1}>=0\)

=>x-1<0

=>x<1

1) ĐKXĐ: \(x\notin\left\{0;1\right\}\)

2) Ta có: \(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(1-\dfrac{3-\sqrt{x}}{\sqrt{x}+1}\right)\)

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

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

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