a: Thay x=1 vào A, ta được:
\(A=\dfrac{1+2}{1-2}=\dfrac{3}{-1}=-3\)
b: \(B=\dfrac{1}{\sqrt{x}-2}+\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{2\sqrt{x}}{4-x}\)
\(=\dfrac{1}{\sqrt{x}-2}+\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}+2+\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{-\sqrt{x}+2+x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{x-2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt[]{x}}{\sqrt{x}+2}\)
c: \(A\cdot B>=0\)
=>\(\dfrac{\sqrt{x}}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}-2}>=0\)
=>\(\dfrac{\sqrt{x}}{\sqrt{x}-2}>=0\)
mà \(\sqrt{x}>=0\) với mọi x thỏa mãn ĐKXĐ
nên \(\sqrt{x}-2>0\)
=>x>4
