[tex]\large A=\dfrac{\sin 4x+\sin 10x-\sin 6x}{\cos 2x+1-2\sin ^24x}=\dfrac{\sin 4x+(\sin 10x-\sin 6x)}{\cos 2x+1+\cos 8x-1}=\dfrac{2\sin 2x(\cos 8x+ \cos 2x)}{\cos 8x+ \cos 2x}=2 \sin 2x[/tex]
[tex]\large \sin x+ \cos x=\dfrac{1}{5}\Leftrightarrow (\sin x+ \cos x)^2=\dfrac{1}{25}\Leftrightarrow \sin^2x+\cos^2x+2\sin x\cos x=\dfrac{1}{25}\Leftrightarrow 2\sin x\cos x = \dfrac{1}{25}-1=-\dfrac{24}{25}\Leftrightarrow \sin2x=-\dfrac{24}{25}[/tex]
[tex]\large \Rightarrow A=-\dfrac{48}{25}[/tex]