$-6+\log3\left(x-8\right)=-4$
$-9+4-1$
$6x^2+3a^2=12$
$\int\frac{\left(4x+6\right)}{\left(x^2+3x-8\right)}dx$
$-x^6+2x^5y-3x^2y^4-xy^5$
$\frac{dy}{dx}=0.08x^2$
$2^3\:\cdot3^3\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$
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