$\left(4+2\right)^2-12$
$x^2-4x+5<0$
$\lim_{x\to\infty}\left(\frac{1+\sec\left(x\right)}{\tan\left(x\right)}\right)$
$3x^3+2x^3$
$9-\left(9-7\right)$
$0.002613532\left(x-1.7\right)^2$
$-\:32\:-\:\left[19\:-\:\left(24\:-\:46\right)\right]$
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