$x^2=e^c\sqrt{t-1}$
$\frac{\left(\frac{1}{x}-\frac{1}{2}\right)}{x-2}$
$\int\frac{\left(x+1\right)}{\left(x+4\right)\left(x-4\right)}dx$
$\frac{dy}{dx}\:y=\left(x^2+1\right)^3$
$-5.55-8.55$
$x^2+3x-6=0$
$cos\:x\left(\frac{sen\:x}{cos\:x}+\frac{cos\:x}{senx}\right)$
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