a: \(A=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{8\sqrt{x}}{x-1}\right):\left(\dfrac{-x+\sqrt{x}-3}{x-1}-\dfrac{1}{\sqrt{x}-1}\right)\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2-8\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\dfrac{-x+\sqrt{x}-3-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-8\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{-x-4}\)
\(=\dfrac{-4\sqrt{x}}{-x-4}=\dfrac{4\sqrt{x}}{x+4}\)
b: \(A=\dfrac{4}{5}\)
=>\(\dfrac{4\sqrt{x}}{x+4}=\dfrac{4}{5}\)
=>\(\dfrac{\sqrt{x}}{x+4}=\dfrac{1}{5}\)
=>\(x+4=5\sqrt{x}\)
=>\(x-5\sqrt{x}+4=0\)
=>\(\left(\sqrt{x}-1\right)\left(\sqrt{x}-4\right)=0\)
=>\(\left[{}\begin{matrix}x=1\left(loại\right)\\x=16\left(nhận\right)\end{matrix}\right.\)
