$\frac{7x^2+4}{x\left(7x-31\right)}=\frac{\left(7x+4\right)}{7x-31}$
$\left(2^x+2^{-x}\right)^2$
$-y-10y$
$2x^2-3x+2\ge0$
$\left(x^2+4x+5\right)+\left(3x^2-2x-6\right)$
$-2y^2-15y-27$
$a^2\:mas\:ab$
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