$0<\frac{9x^2+12x+4}{4}$
$2\:\left(x\:+\:6\right)\:+\:3x\:+\:4$
$2\int\frac{1}{\left(4x^2+1\right)\left(2x+1\right)}dx$
$a^5+9a^3$
$\left(x-3\right)\left(x^2+2x+7\right)$
$\frac{3x^4}{12x^3}$
$\sin x=\frac{15}{17}$
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