ĐKXĐ: \(a>0,a\ne 0\)
a/ \(Q=(\dfrac{\sqrt a}{2}-\dfrac{1}{2\sqrt a})^2.(\dfrac{\sqrt a-1}{\sqrt a+1}-\dfrac{\sqrt a+1}{\sqrt a-1})\\=(\dfrac{a}{2\sqrt a}-\dfrac{1}{2\sqrt a})^2.(\dfrac{(\sqrt a-1)^2}{(\sqrt a+1)(\sqrt a-1)}-\dfrac{(\sqrt a+1)^2}{(\sqrt a+1)(\sqrt a-1)})\\=\dfrac{(a-1)^2}{4a}.\dfrac{a-2\sqrt a+1-a-2\sqrt a-1}{a-1}\\=\dfrac{(a-1)^2}{4a}.\dfrac{-4\sqrt a}{a-1}\\=\dfrac{1-a}{\sqrt a}\)
b/ \(Q<0\\→\dfrac{1-a}{\sqrt a}<0\)
mà \(\sqrt a>0\) (vì \(a>0\) )
\(→1-a<0\\↔a>1\)
c/ \(Q=-2\\→\dfrac{1-a}{\sqrt a}=-2\\→1-a=-2\sqrt a\\↔-a+2\sqrt a+1=0\\↔a-2\sqrt a-1=0\\↔a-2\sqrt a+1-2=0\\↔(\sqrt a-1)^2-2=0\\↔(\sqrt a-1-\sqrt 2)(\sqrt a-1+\sqrt 2)=0\\↔\left[\begin{array}{1}\sqrt a-1-\sqrt 2=0\\a-1+\sqrt 2=0\end{array}\right.\\↔\left[\begin{array}{1}\sqrt a=1+\sqrt 2\\\sqrt a=1-\sqrt 2\end{array}\right.\\↔\left[\begin{array}{1}a=3+2\sqrt 2(TM)\\a=3-2\sqrt 2(KTM)\end{array}\right.\\↔a=3+2\sqrt 2\)
Vậy \(a=3+2\sqrt 2\) thì \(Q=-2\)
d/ \(T=Q\sqrt a\\=\dfrac{1-a}{\sqrt a}.\sqrt a\\=1-a\)
mà \(a>0\)
\(→1-a<1\)
Vậy \(T<1\)
