\(P=\left[\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{x+\sqrt{x}}{x-1}\right]:\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}\right)\)
<=> \(P=\left[\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]:\dfrac{\sqrt{x}-1+\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
<=> \(P=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
<=> \(P=\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}-1}\cdot\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{2\sqrt{x}}\)
<=> \(P=\dfrac{\sqrt{x}+1}{2\sqrt{x}}\)
Khi \(x=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\) <=> \(\sqrt{x}=\sqrt{2}-1\)
=> \(P=\dfrac{\sqrt{2}-1+1}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{2}}{2\sqrt{2}-2}=\dfrac{2+\sqrt{2}}{2}\)
a: Ta có: \(P=\left(\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{x+\sqrt{x}}{x-1}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}\right)\)
\(=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}-1+\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{1}{\sqrt{x}-1}\cdot\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{2\sqrt{x}}\)
\(=\dfrac{\sqrt{x}+1}{2\sqrt{x}}\)
