$\frac{x^4+4x^3-4x^2+5}{x-2}$
$-2-2\cdot3$
$6\:\frac{3}{2}$
$\left(\tan^2a+1\right)\sin^2a=\tan^2a$
$\left(3+i\right)^4$
$n+12=25$
$\left(\frac{3x^2-1}{2x-1}\right)^3$
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