$-4x+6y-6y+3x$
$4+x>5-2x$
$\frac{1}{3}\cdot\ln\left(3e^x+e^6-3\right)$
$-\left|-20\right|$
$-2x^2+12x-8y-18=0$
$\left(+5a^2b^3\right)\left(-6b^4\right)$
$\sqrt{3+y}+\sqrt{y}=3$
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