$\:\left(4x^2-5y^4\right)\left(4x^2+2y^4\right)$
$x+3<5$
$3x^3+9x^2+9x+3$
$x^2+9x-5$
$b\cdot b^x$
$\frac{x}{6}-\frac{x}{8}=1$
$4a^2+4ax+x^2$
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