$3x-9<6$
$\frac{x^7+x^5+1}{x^3+1}$
$\frac{d}{dx}\left(sin\left(y\right)-4cos\left(x\right)=ln\left(x^2+y^2\right)\right)$
$-27\cdot33$
$-\frac{2}{3}+2.4x\sqrt[3]{8}-\frac{5}{6}$
$-3x+4;\:x=1$
$\frac{6x^3-3x^2-8}{2x+3}$
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