$\tan\left(x\right)=\frac{\sin\left(0\right)}{\cos\left(0\right)}$
$\frac{x^2+6x+2}{x^2-3}$
$3x\:+\:2x\:+\:x\:-\:5x\:+\:\left(3x\:-\:2x\:-\:x\right)\:=\:\:$
$5x<3x-5$
$\int\left(\frac{12}{\left(4x-9\right)^4}\right)dx$
$12z^2\:-\:16z\:+\:4$
$y'\:=2\left(y-1\right)$
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