$\:f\left(x\right)=\left(4x\right)^{ln4x}$
$\lim_{x\to+\infty}\left(\frac{3x+4}{\sqrt{2x^2-5}}\right)$
$3f\:+\:f\:+\:2f\:-\:8f\:+\:4f\:+\:3f$
$\left(-7-4i\right)+\left(7-5i\right)-\left(-3+2i\right)$
$\left(2^1\right)\left(2^m\right)$
$\left(9a^3b^3x\:+\:5\:\right)^3$
$-5x^2+10x+15$
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