9:
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x< >1\end{matrix}\right.\)
b: \(A=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{6\sqrt{x}}{x-1}\right):\dfrac{x+2\sqrt{x}}{x-1}\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2+6\sqrt{x}}{x-1}\cdot\dfrac{x-1}{x+2\sqrt{x}}\)
\(=\dfrac{x+2\sqrt{x}+1-x+2\sqrt{x}-1+6\sqrt{x}}{x+2\sqrt{x}}\)
\(=\dfrac{10\sqrt{x}}{x+2\sqrt{x}}=\dfrac{10}{\sqrt{x}+2}\)
c: \(x=12-8\sqrt{2}=4\left(3-2\sqrt{2}\right)=4\left(\sqrt{2}-1\right)^2\)
\(=\left(2\sqrt{2}-2\right)^2\)
Thay \(x=\left(2\sqrt{2}-2\right)^2\) thì \(A=\dfrac{10}{2\sqrt{2}-2+2}=\dfrac{10}{2\sqrt{2}}=\dfrac{5}{\sqrt{2}}=\dfrac{5\sqrt{2}}{2}\)
c: Để A là số nguyên dương thì \(\sqrt{x}+2\inƯ\left(10\right)\)
=>\(\sqrt{x}+2\in\left\{5;10\right\}\)(Vì căn x+2>2 với mọi x thỏa mãn ĐKXĐ)
=>\(\sqrt{x}\in\left\{3;8\right\}\)
=>\(x\in\left\{9;64\right\}\)
