$2y^2=-3x$
$f\left(x\right)=\sqrt{\frac{\left(x^2-4\right)}{\left(x^2+4\right)}}$
$convert\left(111\right)base\:2$
$\left(3x^5\:+\:2y^3\right)\left(\:3x^5\:+\:2y^3\right)$
$-2^2-8x-2+1$
$\left(-8\right)^5$
$x^2-x+8\ge2x^2-2$
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