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$\frac{4x}{-8}$
$\lim_{x\to+\infty}\left(\sqrt{x}-\ln\left(x\right)^2\right)$
$-8\left(2\right)+\left(-4\right)$
$x^2-4x-3$
$\left(\frac{\left[\sqrt[4]{x+1}-\sqrt[4]{1}\right]-\left[\sqrt[3]{x+1}-\sqrt[3]{1}\right]}{x}\right)$
$\left(3y^2+2x^2\right)^2$
$3x^2-300$
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