$\left(2x^3+3\right)\left(2x^3+7\right)$
$\lim_{x\to4}\left(\frac{x}{\sqrt{x^2-9}}\right)$
$13x^2-15x^2+3x^2-4x+5x^2-11x+x^2+9x-4x^2$
$\left(9c^7+0.3\right)\left(9c^7-0.3\right)$
$\left(-6x^2\right):\left(x^2-x+3\right)$
$\left(4-2\sqrt{5}z\right)$
$\frac{1}{\:2}x+\frac{3}{4}x+\frac{4}{3}x+\frac{1}{6}x$
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