Question

B1: Cho 
\(M=\dfrac{x-2}{x+2\sqrt{x}}-\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\)     với x>0

a, Rút gọn M 

b, so sánh M với 1

Answer 1

Answer 2

a) Ta có: \(M=\dfrac{x-2}{x+2\sqrt{x}}-\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\)

\(=\dfrac{x-2}{\sqrt{x}\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x-2-\sqrt{x}-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\)

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)

b) Ta có: M-1

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}}\)

\(=\dfrac{-2}{\sqrt{x}}< 0\forall x\) thỏa mãn ĐKXĐ

hay M<1