$5x^2\:-\:6xy\:+\:y^2\:por\:4x\:-\:11y$
$\int\frac{1}{\sqrt{x}\cdot\left(x+4\right)}dx$
$5a^4b\cdot5a^2b^2$
$cos\left(-15\right)cos\left(75\right)+sin\left(-15\right)sin\left(75\right)$
$-5z-10z-10z$
$6x+5=17$
$\frac{1}{\sqrt{2+x}}$
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