$\int\left(\left(x^3+3\right)^4\left(3x^2\right)\right)dx$
$\frac{7}{2}\left(2x^3+9x^2+9x+7\right)$
$29-17-19+21-13-7+9+11-6-1$
$\left(\frac{1}{16}m+\frac{2}{5}n\right)\left(\frac{1}{4}m-\frac{2}{5}n\right)$
$10a^2+15a$
$345.67907\cdot1000$
$\sqrt{x+1}=9$
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