$\frac{6x+9}{5}\:\le2x-4$
$\left(1+\frac{1}{x}\right)^x$
$6\:x\:\frac{3}{2}$
$0546\cdot75$
$\int\frac{\left(x^3-4x^2-6x+2\right)}{x^2\left(x+1\right)\left(x+2\right)}dx$
$3\left(2-7x\right)>7\left(1-2x\right)$
$\frac{-6x-2-31x}{x-4}$
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