$\sqrt{16a^5}b^4c^3$
$\lim_{x\to8}\left(\frac{\sqrt{7+\sqrt[3]{x-3}}}{x-8}\right)$
$6z^3+12z^2+10z$
$\left(8x\cos\left(2x\right)\right)dx+\left(2y^5+5\right)dy=0$
$\lim_{x\to0}\left(\frac{5.\sin\left(x\right)}{x+3x^2}\right)$
$-3\left(2x\right)^2\left(-2\right)$
$5x+4=7x-8$
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