a: \(P=\left(\dfrac{1}{\sqrt{x}-2}+\dfrac{5\sqrt{x}-4}{2\sqrt{x}-x}\right):\left(\dfrac{2+\sqrt{x}}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\)
\(=\left(\dfrac{1}{\sqrt{x}-2}-\dfrac{5\sqrt{x}-4}{x-2\sqrt{x}}\right):\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-x}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\)
\(=\dfrac{\sqrt{x}-5\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{x-4-x}\)
\(=\dfrac{-4\left(\sqrt{x}-1\right)}{-4}=\sqrt{x}-1\)
b: Khi \(x=\dfrac{3-\sqrt{5}}{2}=\dfrac{6-2\sqrt{5}}{4}=\left(\dfrac{\sqrt{5}-1}{2}\right)^2\) thì
\(P=\sqrt{\left(\dfrac{\sqrt{5}-1}{2}\right)^2}-1=\dfrac{\sqrt{5}-1}{2}-1=\dfrac{\sqrt{5}-3}{2}\)
c: x-2P-1<0
=>\(x-2\sqrt{x}+2-1< =0\)
=>\(x-2\sqrt{x}+1< =0\)
=>\(\left(\sqrt{x}-1\right)^2< =0\)
=>\(\sqrt{x}-1=0\)
=>x=1(nhận)
d: Để 5/P nguyên thì \(5⋮\sqrt{x}-1\)
=>\(\sqrt{x}-1\in\left\{1;-1;5;-5\right\}\)
=>\(\sqrt{x}\in\left\{2;0;6;-4\right\}\)
=>\(\sqrt{x}\in\left\{2;6\right\}\)(theo đkxđ thì căn x>0)
=>\(x\in\left\{4;36\right\}\)
kết hợp ĐKXĐ, ta được: x=36
