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

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

\(=\dfrac{\left(1-x\right)^2}{2\sqrt{x}}.\dfrac{-4\sqrt{x}}{-\left(1-x\right)}\)

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

\(=2\sqrt{x}-2x\sqrt{x}\)

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

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

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

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

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

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

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

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

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

\(A_1=\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)

\(A_2=\left[\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{\left(x+1\right)\left(\sqrt{x}-1\right)}\right]:\dfrac{x-\sqrt{x}+1}{x+1}\\ A_2=\dfrac{x-1}{\left(\sqrt{x}-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{x-\sqrt{x}+1}\\ A_2=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+1}{x-\sqrt{x}+1}\)

b) Ta có: \(4x^2+x-5=0\)

\(\Leftrightarrow4x^2-4x+5x-5=0\)

\(\Leftrightarrow4x\left(x-1\right)+5\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(4x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-\dfrac{5}{4}\left(loại\right)\end{matrix}\right.\)

Thay x=1 vào biểu thức \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}\), ta được:

\(B=\dfrac{\sqrt{1}-1}{\sqrt{1}}=0\)

Vậy: Khi \(4x^2+x-5=0\) thì B=0

\(=\dfrac{3\sqrt{x}-x+2x}{9-x}:\dfrac{\sqrt{x}-1-2\sqrt{x}+6}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{-\sqrt{x}+5}\)

\(=\dfrac{x}{\sqrt{x}-5}\)