a: Thay x=25 vào A, ta được:
\(A=\dfrac{2}{5-1}=\dfrac{2}{4}=\dfrac{1}{2}\)
b. \(P=\dfrac{2}{\sqrt{x}-1}+\dfrac{2}{\sqrt{x}+1}-\dfrac{5-\sqrt{x}}{x-1}\)
\(\Rightarrow P=\dfrac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{5-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow P=\dfrac{2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{5-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow P=\dfrac{2\sqrt{x}+2+2\sqrt{x}-2-5+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow P=\dfrac{5\sqrt{x}-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow P=\dfrac{5\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(\Rightarrow P=\dfrac{5}{\sqrt{x}+1}\)
c. P>2\(\Rightarrow\dfrac{5}{\sqrt{x}+1}>2\\ \Rightarrow5>2\sqrt{x}+2\\ \Rightarrow3>2\sqrt{x}\\ \Rightarrow\dfrac{3}{2}>\sqrt{x}\\ \Rightarrow x< \dfrac{9}{4}\)
