$\int\left(2^{2x}\cdot5x\right)$
$\frac{x^5}{3x}$
$-16+m^4$
$8\left(-2\right)\cdot4\left(-2\right)$
$\lim_{x\to\pi}\left(\frac{1+\cos\left(x\right)}{\sin\left(2x\right)}\right)$
$\left|27-16\right|$
$\frac{8\cos^2x+10\cos x+2}{2\cos^2x+4\cos x+2}$
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