$\frac{3}{4}x\cdot6x^2\cdot\left(-2x^3\right)$
$b\frac{3}{4}+b-\frac{1}{3}$
$-7+1\cdot16-1\cdot1^3-1\cdot5^2$
$8-2\left(+3\right)+2\left(+1\right)$
$\lim_{x\to0}\left(\frac{1+x-sen\left(x\right)}{x^3}\right)$
$xy'+2y=x^3$
$ax-bx$
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