$3-5x>10-7x$
$\left(3x^4+3\right)^5\:\cdot\left(9x\right)^{7x}$
$\frac{\infty^3}{\infty^4}$
$2x^3+4x^4+5x-4x^3$
$x^5+x^3\cdot y^2=4$
$\int5\left(-3x+7\right)^{\frac{3}{7}}dx$
$y'+\frac{y}{\:x}-3x=0$
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