$\left(1+4i\right)\left(3+6i\right)$
$15-x=13$
$-x^2+6x+7=0$
$\lim_{x\to7}\left(\frac{7-x+ln\left(x-6\right)}{x^3-12\cdot x^2+21\cdot x+98}\right)$
$-4\left(1-5\right)-\left(4^2\right)$
$\left(\frac{3}{7}a^2-15\right)\left(\frac{3}{7}a^2+8\right)$
$-4+y+4y-5x+2x-3$
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