Question

M=\(\left(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{x}\)

tính Min M

Answer 1

ĐKXĐ: \(x>0;x\ne1\)

\(M=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}+1}{x}\)

\(=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right).\dfrac{x}{\sqrt{x}+1}=\dfrac{\left(x-1\right)}{\sqrt{x}}.\dfrac{x}{\sqrt{x}+1}\)

\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)x}{\sqrt{x}\left(\sqrt{x}+1\right)}=\sqrt{x}\left(\sqrt{x}-1\right)\)

\(=x-\sqrt{x}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

\(M_{min}=-\dfrac{1}{4}\) khi \(x=\dfrac{1}{4}\)