$\frac{1}{5}\left(x^2-4x+3\right)<0$
$-4.\:-13.-5\:$
$\frac{\cos^2x-\sin^2x}{\sin x\:+\cos x}=\cos x\:-\:\sin x$
$3\left(v+6\right)-v\left(6-3v\right)$
$1-\left(\sin\left(a\right)+\cos\left(a\right)\right)^2$
$-x^2+8x-13$
$\frac{t^2}{t-8}$
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