$\left(2x+3yc\right)^2$
$\frac{3}{\frac{3}{0}}$
$25a^2+20a+9$
$2\left(x+1\right)+4>5\left(x-2\right)+7$
$a^2b-7ab-6ab-5a^2b+3ab$
$\frac{0}{\left(-10\right)}$
$\left(\frac{1}{2}x^3-x^2\right)\cdot\left(8x^3+\frac{1}{2}\right)$
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