\(A=\dfrac{-1}{\sqrt{a}+1}\) hay \(A=\dfrac{-1}{\sqrt{a}}+1\)
Bài này đơn giản mà :v
Thay \(a=\dfrac{4}{9}\) vào biểu thức A ta được :
\(A=\dfrac{-1}{\sqrt{a}+1}=\dfrac{-1}{\sqrt{\dfrac{4}{9}}+1}=\dfrac{-1}{\dfrac{2}{3}+1}-\dfrac{3}{5}\)
\(\left|A\right|=\dfrac{1}{2}\Leftrightarrow\left|\dfrac{-1}{\sqrt{a}+1}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-1}{\sqrt{a}+1}=\dfrac{1}{2}\left(1\right)\\\dfrac{1}{\sqrt{a}+1}=\dfrac{1}{2}\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{a}+1=-2\) ( KTM )
\(\left(2\right)\Leftrightarrow\sqrt{a}+1=2\Leftrightarrow a=1\) ( KTM )