$\frac{x-11}{10}=\frac{x-10}{8}$
$\lim_{x\to\infty}\left(\frac{4x-1}{4x+1}\right)^{\left(4x\right)}$
$b^2+2b-c^2-2c$
$-54m^2n-36mn^2-8n^3$
$x^2-5x\ge\:-8$
$\left(\frac{1}{2}x+\frac{7}{8}\right)\left(\frac{1}{5}x+\frac{2}{3}\right)$
$\:\left(y^6+3x\right)^2$
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