Bài 3:
a) \(A=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)^2\left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}-\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\right)=\left(\dfrac{a-1}{2\sqrt{a}}\right)^2\left(\dfrac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)=\dfrac{\left(a-1\right)^2}{4a}.\dfrac{-4\sqrt{a}}{a-1}=\dfrac{1-a}{\sqrt{a}}\)b) \(A< 0\Leftrightarrow\dfrac{1-a}{\sqrt{a}}< 0\Leftrightarrow1-a< 0\Leftrightarrow a>1\)( do \(\sqrt{a}>0\))
Bài 3:
a: Ta có: \(A=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)^2\cdot\left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}-\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
\(=\dfrac{\left(a-1\right)^2}{4a}\cdot\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\dfrac{-4\sqrt{a}\left(a-1\right)}{4a}\)
\(=\dfrac{-a+1}{\sqrt{a}}\)
b: Để A<0 thì \(-a+1< 0\)
\(\Leftrightarrow-a< -1\)
hay a>1
