$\sec\left(a\right)+5=7$
$\frac{3}{4z+2}=\frac{1}{z+2}$
$x^2-4x+1^2$
$\frac{1}{sen\:\left(a\right)\:cos\:\left(a\right)\:}-\frac{\cos\left(a\right)}{sen\left(a\right)}=\tan\left(a\right)$
$\left(2x^3+5b^2\right)\left(2x^3-5b^2\right)$
$x^2\cdot y'=y^2$
$2y^2\:+\:2y^2$
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