a) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)

Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right):\dfrac{2\sqrt{x}}{x-4}\)

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

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

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

\(=\sqrt{x}\)

b) Để P>4 thì \(\sqrt{x}>4\)

hay x>16

Kết hợp ĐKXĐ, ta được: x>16

Vậy: Khi x>16 thì P>4

\(a,ĐK:x>0;x\ne9\\ b,A=\dfrac{\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\\ A=\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{2}{\sqrt{x}+3}\\ c,A>\dfrac{2}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+3}-\dfrac{2}{5}>0\\ \Leftrightarrow\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{5}>0\\ \Leftrightarrow\dfrac{2-\sqrt{x}}{5\left(\sqrt{x}+3\right)}>0\\ \Leftrightarrow2-\sqrt{x}>0\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow\sqrt{x}< 2\Leftrightarrow0< x< 4\)

a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)

b: Ta có: \(D=\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{5\sqrt{x}+5}{x-4}\right)\cdot\dfrac{x-4}{\sqrt{x}}\)

\(=\dfrac{x+4\sqrt{x}+4-x+4\sqrt{x}-5\sqrt{x}-5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{x-4}{\sqrt{x}}\)

\(=\dfrac{3\sqrt{x}-1}{\sqrt{x}}\)

a) ĐKXĐ: \(x\ge0;x\ne1\)

b) \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{2}{\sqrt{x}+1}\left(x\ge0;x\ne1\right)\\ P=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{2}\\ P=\dfrac{x-\sqrt{x}-x-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{2}\\ P=\dfrac{-2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{2}\\ P=\dfrac{-\sqrt{x}}{\sqrt{x}-1}\)

Giúp mình với

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

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

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

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

a) ĐKXĐ: 

\(\left\{{}\begin{matrix}\sqrt{x}-2>0\\\sqrt{x}+2>0\\\sqrt{4x}>0\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}\sqrt{x}>2\\\sqrt{x}>-2\\2\sqrt{x}>0\end{matrix}\right.\\\rightarrow \left\{{}\begin{matrix}x>\sqrt{2}\\x>-\sqrt{2}\\x>0\end{matrix}\right.\\ \rightarrow x>\sqrt{2}\)

Vậy \(x>\sqrt{2}\)

b) 

\(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right).\dfrac{x-4}{\sqrt{4x}}\\ =\left[\dfrac{\sqrt{x}.\left(\sqrt{x}+2\right)+\sqrt{x}.\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right].\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\sqrt{x}}\\ =\dfrac{x+2\sqrt{x}+x-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\sqrt{x}}\\ =\dfrac{2x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\sqrt{x}}\\ =\dfrac{2x}{2\sqrt{x}}=\dfrac{x}{\sqrt{x}}=\dfrac{\sqrt{x}.\sqrt{x}}{\sqrt{x}}=\sqrt{x}\)

Vậy \(M=\sqrt{x}\)

a) ĐKXĐ:

\(\left\{{}\begin{matrix}\sqrt{x}-2>0\\\sqrt{x}+2>0\\\sqrt{4x}>0\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}\sqrt{x}>2\\\sqrt{x}>-2\\2\sqrt{x}>0\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}x>4\\x>-4\\x>0\end{matrix}\right.\\ \rightarrow x>4\)

Vậy \(x>4\)

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:

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