$\frac{7}{7^{-3}}$
$x2+10x+16=0$
$x+1=2$
$\left(\left(x-2\right)\cdot\left(x-1\right)\right)\cdot\left(\left(x+2\right)\cdot\left(x+1\right)\right)$
$2,23-3$
$\lim_{x\to\infty}\left(\frac{7+x}{x}\right)^{5x}$
$\frac{5+3}{4}$
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