$f\left(t\right)=\left(t^4-1\right)^3\left(t^3+1\right)^4$
$\lim_{x\to0}\left(\frac{\sin\left(x\right)}{x^2}.\left(tan\left(4x-tan2x\right)\right)\right)$
$\frac{x^61}{x-1}$
$-2\ln x+1=1$
$x^2+3x-72=0$
$\frac{6}{x^2-4x+3}$
$-7x^2\:-\:x\:+\:6x\:+\:x^2$
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