\(\lim_{x\to 0} \dfrac{\sqrt[5]{x+1}-1}{x}=\lim_{x\to 0}\dfrac{1}{(\sqrt[5]{x+1})^4+(\sqrt[5]{x+1})^3+(\sqrt[5]{x+1})^2+\sqrt[5]{x+1}+1}=\dfrac{1}{5}\)

\(sin 2x-(2sin^2 x-sin2x-2sinx-1/2.\sin 2x+\cos^2x+\cos x-3\sin x-3\cos x+3)=0\)

\(5\sin x.\cos x+5\sin x+2\cos x-\sin^2x-4=0\)

\(\cos x(5\sin x+2)=\sin^2x-5\sin x+4=(\sin x-1)(\sin x -4)\)

Bình phương 2 vế suy ra

\((1-\sin^2 x)(5\sin x+2)^2=(1-\sin x)^2(\sin x-4)^2\)

TH1: \(\sin x=1\)

TH 2: \((1+\sin x)(5\sin x+2)^2=(1-\sin x)(\sin x-4)^2\)