$f\left(x\right)=\left(\frac{x^3}{1+7x}\right)^3$
$\frac{12}{0.37}$
$\lim_{x\to3}\left(4x^3-5x^2+x-6\right)$
$-4-5+1-1$
$3+12\cdot19-10$
$3x+5x\left(2x-1\right)$
$x^2-16x-17=0$
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