a: \(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right)\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\left(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right)\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot2\)
\(=\dfrac{-\sqrt{x}+1+x-\sqrt{x}}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{2}{x+\sqrt{x}+1}=\dfrac{2}{x+\sqrt{x}+1}\)
b: Thay x=4 vào A, ta được:
\(A=\dfrac{2}{4+2+1}=\dfrac{2}{7}\)
c: \(x+\sqrt{x}+1>=1\forall x\) thỏa mãn ĐKXĐ
=>\(A=\dfrac{2}{x+\sqrt{x}+1}< =\dfrac{2}{1}=2\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi x=0
