\(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{x-\sqrt{x}-1}{x-2\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{x-5}{x-\sqrt{x}-2}\right)\)
DKXD: \(x\ne4;x>0\)
\(=\left(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{x-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}\right):\left(\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}-\dfrac{x-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-2\right)}.\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b) minh kh hieu de lam sorry
c)
Thay \(x=6-2\sqrt{5}\) vao P:
\(\Leftrightarrow P=\dfrac{\sqrt{6-2\sqrt{5}}+1}{\sqrt{6-2\sqrt{5}}}=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}+1}{\sqrt{\left(\sqrt{5}-1\right)^2}}=\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{4}\)
a: Ta có: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{x-\sqrt{x}-1}{x-2\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{x-5}{x-\sqrt{x}-2}\right)\)
\(=\dfrac{x-x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-2\right)}:\dfrac{x-4-x+5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{1}\)
\(=\dfrac{x+2\sqrt{x}+1}{\sqrt{x}}\)
