$\left(-2x^{-5}y^{-1}\right)^0.x^4y^2$
$\left(x\right)^{12}\frac{x^7}{\left(1-3x\right)^5}$
$\left(1+a^2\right)^3$
$\left(\frac{2}{a}+a^{-1}\right)+\left(\frac{a^2}{2}+\frac{1}{5a^{-2}}\right)$
$y+x^2+6x=1$
$\left|16^5\right|^0$
$448\cdot\frac{1}{8}$
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