a: Thay x=25 vào A, ta được:
\(A=\dfrac{5-2}{5}=\dfrac{3}{5}\)
b: \(B=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{2x}{x-9}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{2x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-2x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-2x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
c: \(P=A\cdot B=\dfrac{\sqrt{x}}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}}=\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\)
Để P<0 thì \(\sqrt{x}-2< 0\)
=>\(\sqrt{x}< 2\)
=>0<=x<4
mà x nguyên và x>0; x<>9
nên \(x\in\left\{1;2;3\right\}\)
