$\left(2-1\right)^2\left(2^5+2^3+2^2+1\right)$
$\lim_{x\to\infty}\frac{x\arcsin\left(0\right)}{\arctan^2\left(x\right)}$
$4\left(4b+5\right)$
$\left(4a^2+6\right)\cdot\left(a+4\right)$
$x-7<-9x+13$
$x\tan^2ydx+\left(1+\sqrt{x}\right)dy=0$
$-18+6x+2-x$
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