$3x^2+3y^2=-6$
$\frac{2x+3}{x+1}>10$
$\left(\frac{4}{9}x^6+\frac{1}{25}y^7\right)\left(\frac{4}{9}x^6+\frac{1}{25}y^7\right)$
$\sqrt{\frac{a+1}{a-1}}$
$\frac{d}{dx}x^2-y^2-7=0$
$\left(1-2n\right)\left(1+2n\right)+4\left(n-1\right)\left(n+2\right)$
$9x^2+16y^2=49$
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