$12\le6-2x$
$-10x+3<\frac{1}{5}x+3$
$sen3x-sen6x=0$
$\left(\sqrt[3]{6x^3y^2}\right)\left(\sqrt[3]{2x^2y^5}\right)$
$x^4+2x^3+x^2$
$\:\frac{x^4-2x^2-2x+35}{x+3}\:\:$
$x^2-15x\:+8\:=0$
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