$41n\left(x-5\right)=-8$
$\left(4x-7\right).\left(3x-4\right)$
$\left(\frac{2}{3}\right)^{-4}\left(\frac{3}{2}\right)^2.2^4.3^5$
$4x+24x^3$
$\frac{\left(x+2\right)}{\left(x+2\right)^2}\cdot\frac{\left(x^2-4\right)}{x}$
$-3+5-6+4-1-4$
$-2\left(8-1\right)$
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