$6+v+45$
$\left(5+10^4\right)-\left(3+10^2\right)$
$\left(-7\right)\left(8\right)\left(0\right)\left(3\right)$
$\left(4a^4-9b^4\right)^2$
$\frac{5x^4y^4-16x^3y^5+36x^5y^3\:}{8x^3y^3}$
$\frac{2x^4-8x^2-2x+4}{x-4}$
$3-sec^2x$
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