$4-\frac{3}{x}=6-\frac{5}{x}$
$x^4+7x^3+17x^2+17x+6$
$\left(\sqrt[3]{7+\sqrt{22}}\right)\left(\sqrt[3]{7-\sqrt{22}}\right)$
$-x+2-y+4x-4+7y$
$\lim_{x\to a}\left(\frac{6^x-2^x}{x}\right)$
$y=x^3-2x^2+x$
$\lim_{n\to0}\left(\frac{n^3-3n^2-4n}{n^3+3x^2-4x}\right)^n$
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