a: \(A=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{3}{\sqrt{x}-3}+\dfrac{x+9}{9-x}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{3}{\sqrt{x}-3}-\dfrac{x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+3\left(\sqrt{x}+3\right)-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2x-6\sqrt{x}+3\sqrt{x}+9-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
b: \(A< \dfrac{1}{2}\)
=>\(A-\dfrac{1}{2}< 0\)
=>\(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{1}{2}< 0\)
=>\(\dfrac{2\sqrt{x}-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}< 0\)
=>\(\dfrac{\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}< 0\)
=>\(\sqrt{x}< 3\)
=>0<=x<9
Kết hợp ĐKXĐ, ta được: 0<=x<9
mà x nguyên
nên \(x\in\left\{0;1;2;...;8\right\}\)
c: \(\left|A-\dfrac{2}{5}\right|>A-\dfrac{2}{5}\)
=>\(A-\dfrac{2}{5}< 0\)
=>\(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{2}{5}< 0\)
=>\(\dfrac{5\sqrt{x}-2\left(\sqrt{x}+3\right)}{5\left(\sqrt{x}+3\right)}< 0\)
=>\(5\sqrt{x}-2\left(\sqrt{x}+3\right)< 0\)
=>\(3\sqrt{x}-6< 0\)
=>\(\sqrt{x}< 2\)
=>0<=x<4
