$16+6\left(3x\right)-\left(3x\right)^2$
$5x+24=2x+9$
$\lim_{x\to\frac{\pi}{4}}\left(\sqrt{1+\sin\left(\frac{\pi}{2}\tan\left(x\right)\right)}\right)$
$49a^2b^2-16x^4$
$cos2x=2+4sinx$
$\left|2\right|-\left|-7\right|$
$\sec^{2}\theta-5=-1$
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