x−32+2x+62≥ x4−3x−62\frac{x-3}{2}+\frac{2x+6}{2}\ge\:\frac{x}{4}-\frac{3x-6}{2}2x−3+22x+6≥4x−23x−6
limx→0(4x−6xx)\lim_{x\to0}\left(\frac{4^x-6^x}{x}\right)x→0lim(x4x−6x)
x−4=19x-4=19x−4=19
x3y57⋅x4y57\sqrt[7]{x^3y^5}\cdot\sqrt[7]{x^4y^5}7x3y5⋅7x4y5
(x23)14\left(x^{\frac{2}{3}}\right)^{\frac{1}{4}}(x32)41
−(7x⋅ +11)2-\left(7x\cdot\:+11\right)^2−(7x⋅+11)2
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!