$5x\:-\:4\:+\:x$
$\frac{3}{4}x+\frac{1}{5}x$
$x^2+4x=1$
$\lim_{x\to0}\left(\left(cosh\left(t\right)\right)^{\frac{1}{t^2}}\right)$
$sen\left(a\right)tan\left(a\right)sec\left(a\right)+1=sec^2\left(a\right)$
$\frac{d}{dx}\left(x+6\right)\left(x+4\right)\left(x+9\right)\left(x+2\right)$
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