$x+1=\sqrt{x+3}$
$x^4-5x^3+5x^2+5x-6$
$x^2+8x-48$
$\left(a^4+16\right)\left(a^2+4\right)\left(a+2\right)\left(a-2\right)$
$1x+1$
$\lim_{x\to0}\left(\frac{6\sqrt{5+x}-6\sqrt{5}}{x}\right)$
$17=\frac{m}{12}$
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