$\left(2x-3\right)\left(2x+3\right)-\left(x-4\right)=3x^2+8x-25$
$-3a+4b-6a+81b-11b+31a-b$
$x^2+12x=144$
$tn^{\circ}8=\left(2x-1\right)^{15}$
$18+9+19+-8$
$\frac{d\left[u\right]}{dt}=-k\left[u\right]$
$\frac{x^2+18x+81}{x^3-27}$
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