$A=\dfrac{sinx-2cos3x-sin5x}{cosx+2sinx-cos5x}$

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1. Đáp số

   $A=\dfrac{-\cos3x(\sin 2x+1)}{\sin2x(\sin3x+1)}$

   Lời giải chi tiết

   $A=\dfrac{\sin x-2\cos3x-\sin5x}{\cos x+2\sin 2x-\cos5x}\\ \\
   =\dfrac{(\sin x-\sin5x)-2\cos3x}{(\cos x-\cos5x)+2\sin 2x}\\ \\
   =\dfrac{2\cos \dfrac{x+5x}{2}\sin\dfrac{x-5x}{2}-2\cos3x}{-2\sin\dfrac{x+5x}{2}\sin\dfrac{x-5x}{2}+2\sin 2x}\\ \\
   =\dfrac{2\cos3x\sin(-2x)-2\cos3x}{-2\sin3x\sin(-2x)+2\sin 2x}\\ \\
   =\dfrac{-2\cos3x\sin 2x-2\cos3x}{2\sin3x\sin2x+2\sin 2x}\\ \\
   =\dfrac{-2\cos3x(\sin 2x+1)}{2\sin2x(\sin3x+1)}\\ =\dfrac{-\cos3x(\sin 2x+1)}{\sin2x(\sin3x+1)}$

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