$12a^2+6am$

$\frac{\sin\left(y\right)^2}{\sin\left(y\right)-\cos\left(y\right)^2}=\frac{1}{\sin\left(y\right)}$

$\:y=\frac{1}{x^2},\:x=1,\:x=3$

$x^2\cdot z''\left(x\right)-xy\cdot z''\left(y\right)=z-y$

$\frac{x-6}{3}\le x+4$

$x\cdot x\:+4x-5$

$\left(-7\right)-\left(-3\right)+\left(+5\right)+\left(-9\right)-\left(-6\right)$