$4^{x+3}=\left(\frac{1}{2}\right)^x$
$\left(x+y-3\right)\cdot\left(x-y+5\right)$
$-1\:.\:-2\:.\:-3\:.\:-4\:.\:-5$
$0.30mt\:x\:\frac{2}{6}$
$9\cdot3^2\cdot\left(\frac{6}{2}\right)^2$
$-\left(-3\right)\left(-4\right)$
$\left(\frac{6x+1}{2x-1}\right)^4$
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