\(A=\dfrac{\sqrt{25}+1}{\sqrt{25}-1}=\dfrac{6}{4}=\dfrac{3}{2}\)
\(a,A=\dfrac{5+1}{5-1}=\dfrac{6}{4}=\dfrac{3}{2}\\ b,B=\dfrac{x+1-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\\ c,P=AB=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}=1+\dfrac{1}{\sqrt{x}}>1\left(\dfrac{1}{\sqrt{x}}>0\right)\)
a: Thay x=25 vào A, ta được:
\(A=\dfrac{5+1}{5-1}=\dfrac{6}{4}=\dfrac{3}{2}\)
