a. \(Q=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{1}{x+\sqrt{x}}\right)\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}+1}{x-1}=\dfrac{1}{\sqrt{x}}\)
b. Ta có \(x=3+2\sqrt{2}\Rightarrow\sqrt{x}=\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
Thay \(\sqrt{x}=\sqrt{2}+1\) vào \(Q\), ta được
\(Q=\dfrac{1}{\sqrt{2}+1}\)
c. \(Q< 0\Leftrightarrow\dfrac{1}{\sqrt{x}}< 0\) mà \(1>0\Rightarrow\sqrt{x}< 0\) ( vô lý )
Vậy không có x thỏa ycbt
a: Ta có: \(Q=\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{1}{x+\sqrt{x}}\right)\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)
\(=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{1}{\sqrt{x}}\)