Có: \(A^2=\frac{\left(3x+2y\right)^2}{\left(3x-2y\right)^2}=\frac{9x^2+4y^2+12xy}{9x^2+4y^2-12xy}=\frac{20xy+12xy}{20xy-12xy}=\frac{32xy}{8xy}=4\)
Do: \(2y< 3x< 0\Rightarrow3x-2y>0;3x+2y< 0\)
=>\(A< 0\)
Vậy \(A=-2\)
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