$\left(x+6\right)-\left(2x\\:+7\\:\right)-\\:3x\\:=\\:-9$

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

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

$\left(\frac{5}{3}x^2+\frac{1}{4}y\right)^2$

$5x+8<8x-3$

$\frac{d}{dx}\:\left(\sqrt{x}+1\right).\left(x-\sqrt{2}\right)$

$8x^2y^3\cdot x^4$