a)

\(A=\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\cdot\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)\\ =\left(\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2+4\sqrt{a}\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\cdot\left(\frac{\left(\sqrt{a}\right)^2-1}{\sqrt{a}}\right)\\ =\left(\frac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4a\sqrt{a}-4\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\cdot\left(\frac{a-1}{\sqrt{a}}\right)\)

\(=\frac{4a\sqrt{a}}{a-1}\cdot\frac{a-1}{\sqrt{a}}\\ =4a\)

b)

\(\sqrt{A}>A\Leftrightarrow A-\sqrt{A}< 0\\ \Leftrightarrow4a-\sqrt{4a}< 0\\ \Leftrightarrow4a-2\sqrt{a}< 0\\ \Leftrightarrow2\sqrt{a}\left(2\sqrt{a}-1\right)< 0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2\sqrt{a}< 0\\2\sqrt{a}-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}2\sqrt{a}>0\\2\sqrt{a}-1< 0\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a< 0\left(ktm\right)\\a>\frac{1}{4}\end{matrix}\right.\\\left\{{}\begin{matrix}a>0\\a< \frac{1}{4}\left(tm\right)\end{matrix}\right.\end{matrix}\right.\)

Vậy với \(0< a< \frac{1}{4}\)thì \(\sqrt{A}>A\)

a,đkxđ x>0;x\(\ne\) 1

\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)^2}\)

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

\(=\frac{x-1}{x}\)

b, để P>1/2

\(\frac{x-1}{x}>\frac{1}{2},< =>\frac{x-1}{x}-\frac{1}{2}>0< =>\frac{2\left(x-1\right)-1.x}{2x}< =>\frac{2x-2-x}{2x}< =>x-2>0< =>x>2\)

câu 2 mk ra đáp án là\(\frac{a-\sqrt{a}}{\sqrt{a}}\)mà cũng nghĩ là đúng nhưng câu b tính k đc

ĐK: \(a\ge0\)

a) \(A=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{\sqrt{a}}{a-\sqrt{a}}\right):\frac{\sqrt{a}+1}{a-1}\)

\(A=\left[\frac{a}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]:\frac{\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)

\(A=\frac{a-\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{1}{\sqrt{a}-1}\)

\(A=\frac{\sqrt{a}\left(\sqrt{a}-1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\)

\(A=\sqrt{a}-1\)

b) \(A< 0\)

\(\Leftrightarrow\sqrt{a}-1< 0\)

\(\Leftrightarrow\sqrt{a}< 1\)

\(\Leftrightarrow\left|a\right|< 1\)

\(\Leftrightarrow0\le a< 1\)