$\frac{x}{9}>4$
$5a^2-3a+2-8a^2-_{5a+7}$
$\frac{2x^2+4x+6}{\left(2x+3\right)\left(2x+2\right)}=\frac{17}{35}$
$\left(6^{n+2}\:+\:4b^{m-1}\right)\:\left(6a^{n+2}\:-\:4b^{m-1}\right)$
$x^5-5x^4$
$-1\:-\:\left(\:-3\:\right)$
$5x^2+x^2$
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