$2x^2-2x-8\:=\:-4$
$\left(3x^2-4y^3\right)\left(3x^2+4y^2\right)$
$\frac { - 11 } { - 1 x }$
$4x^2+\:16xy\:+\:16y^2$
$y'-y=xc^{-x}+1$
$12a^3-2a$
$\frac{1}{2\:}x\:-\:\left\{0,5\:-\:\left[0,25x\:-\:\left(\frac{1}{4}\:+\:x\right)\right]\right\}$
Get a preview of step-by-step solutions.
Earn solution credits, which you can redeem for complete step-by-step solutions.
Save your favorite problems.
Become premium to access unlimited solutions, download solutions, discounts and more!