$\lim_{x\to4}\left(\frac{16-x^2}{2\sqrt{x-4}}\right)$
$4\sin\left(x\right)+\sqrt{3}=0$
$\frac { 15 x ^ { 5 } + 21 x ^ { 2 } + 6 } { 3 x + 3 }$
$-8\left(n-7\right)-7n$
$\left(m^2+2n\right)^2$
$16.09+32.4+3.007$
$8cos^2x=16cosx$
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