\(\dfrac{3x-2y}{3x+2y}=\dfrac{\left(3x-2y\right)^2}{\left(3x+2y\right)^2}=\dfrac{9x^2+4y^2-12xy}{9x^2+4y^2+12xy}=\dfrac{1}{4}\)
thay từ đề vào ok
\(\left\{{}\begin{matrix}2y< 3x< 0\\\left\{{}\begin{matrix}3x-2y>0\\3x+2y< 0\end{matrix}\right.\end{matrix}\right.\) \(\left\{{}\begin{matrix}A^2=\dfrac{12xy}{32xy}=\dfrac{12}{32}=\dfrac{3}{8}\\A< 0\Rightarrow A=-\sqrt{\dfrac{3}{8}}\end{matrix}\right.\)
Nhầm:
\(\left\{{}\begin{matrix}2y< 3x< 0\\A< 0\end{matrix}\right.\) \(\left\{{}\begin{matrix}A^2=\dfrac{8xy}{32xy}=\dfrac{8}{32}=\dfrac{1}{4}\\A< 0\Rightarrow A=-\sqrt{\dfrac{1}{4}}=-\dfrac{1}{2}\end{matrix}\right.\)
