$\left(3x^2-4y\right)^3$
$\frac{dy}{dx}+\frac{y}{2}=y^{\left(-2\right)}$
$2x^2-7x=11$
$y2\:+\:7y\:+\:10$
$cosx-1=-cosx$
$\int_1^2\frac{x^2-4x-9}{x^3-6x^2+9x}dx$
$-5-2+4-6-8+2$
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