$\frac{3}{4}+\frac{-2x-1}{2x\left(x+1\right)}=\frac{3x^2-x-2}{4x^2+4x}$

$16\:x\:\:\left(-34\right)\:+\:\left(-34\right)\:x\:\left(-18\right)$

$8\left(-2\right)\left(-3\right)+7\left(-4\right)\left(-3\right)$

$3x^2+132+121$

$\left(4m-5m^2\right)^3$

$16x^2-9=0$

$\tan\left(x\right)\frac{1}{\sin\left(x\right)}=\frac{1}{y}$