$\int_{0}^{\frac{\pi}{4}}\frac{\left(1+\tan x\right)^{3}}{\cos^{2}x}dx$
$\frac{x^2+x}{2x+1}=x-3$
$-3x+10y-z-6x-6y+6z$
$\left(-4\right)\:-\:\left(-7\right)$
$x^2-3x-4<0$
$4x^2+16+16$
$2x-6y+10y+8y+3x\\:-15$
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