$x^2-5x+5$
$x0w$
$16-x^3$
$\sec\left(x\right)^2+4=\tan^2\left(x\right)+5$
$\lim_{x\to\infty}\left(\frac{ln\left(2x\right)}{ln\left(3x\right)}\right)$
$-5,87+48,73-5,4+98,75$
$2a\:+\:3a\:-\:6\:=\:-21\:\:\:\:$
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