$0.005163193046487591^2$
$\left(\frac{x}{3}\right)^4\left(\frac{3}{x}\right)^2$
$x^2+60x+120$
$6>\sqrt{40}$
$\left(5m^5-9n^2\right)\left(5m^5+9n^2\right)$
$-2x-7=1$
$\left(x^4-7x^9\right)^2$
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