$\lim_{x\to1}\left(\frac{ln\left(2x-1\right)}{x-1}\right)$
$16+x=12$
$\left(x^2-\left(1+\frac{1}{x}\right)^2\right)$
$\lim_{x\to\infty}\left(\frac{\left(x^3\cdot e^{-8\cdot x}\right)}{-8}\right)$
$\frac{4x^4+2x^2-6x+5}{x+2}$
$3\cdot\left(-4\right)+\left(52+-4\cdot2\right)-\left(-9.82\right)$
$\frac{4^{-24}}{4^{-3}}$
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