$\left(\frac{dx}{dt}\right)=\left(\frac{\left(1+t\right)}{\left(t^2x^2\right)}\right)$
$b^{3x}+\sqrt{x^2}+1$
$f\:\left(x\right)=x^2+4x+3$
$y=-\frac{1}{4}x^4-\frac{2}{3}x^{-1}$
$x>6-2x$
$y'+3y\:=1$
$15\cdot\left(-9\right)$
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