Question

Answer 1

a: \(P=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

b: Để \(P=\sqrt{x}+\dfrac{5}{2}\) thì \(\dfrac{\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+\dfrac{5}{2}\)

\(\Leftrightarrow2\sqrt{x}+2=2x+5\sqrt{x}\)

\(\Leftrightarrow2x+3\sqrt{x}-2=0\)

\(\Leftrightarrow x=1\)