a: \(A=\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right):\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right):\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}+1}{1}\)
\(=\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}+1}{1}\)
\(=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-2}=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)
b: Đặt P=1:A
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1-3}{\sqrt{x}+1}=1-\dfrac{3}{\sqrt{x}+1}\)
\(\sqrt{x}+1>=1\forall x\) thỏa mãn ĐKXĐ
=>\(\dfrac{3}{\sqrt{x}+1}< =3\forall x\) thỏa mãn ĐKXĐ
=>\(-\dfrac{3}{\sqrt{x}+1}>=-3\forall x\) thỏa mãn ĐKXĐ
=>\(A=-\dfrac{3}{\sqrt{x}+1}+1>=-3+1=-2\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi x=0
