$x^2\:-\:14x\:+\:49\:=\:16$
$\frac{2}{x+2}=\frac{x+1}{3}$
$-7x-18=-11$
$\left(x^4-3y^3\right)^2$
$\sec\left(x\right)-\cos\left(x\right)-\sin\left(x\right)\tan\left(x\right)$
$3\left(x+5\right)-6$
$\frac{dy}{dx}=\frac{6x^2}{\:y\left(2+x^3\right)}$
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