Ta có : M\(^2\)= (\(\dfrac{5x-4y}{5x+4y}\))\(^2\) = \(\dfrac{\left(5x-4y\right)^2}{\left(5x+4y\right)^2}\)= \(\dfrac{25x^2+16y^2-40xy}{25x^2+16y^2+40xy}\)
= \(\dfrac{41xy-40xy}{41xy+40xy}=\dfrac{xy}{81xy}=\dfrac{1}{81}=\left(\dfrac{1}{9}\right)^2\)
Mà 4y < 5x < 0 \(\Rightarrow\)5x - 4y > 0 . 5x +4y < 0 \(\Rightarrow\) M < 0
Vậy M = - \(\dfrac{1}{9}\)