$x^2+4x+5=0$
$\left(y^3+2y^2-3y+1\right)\left(y^2-1\right)$
$\frac{\left(x+5\right)}{3}\ge\frac{\left(4-2x\right)}{-5}$
$5\left(x-1\right)+16\left(2x+3\right)\le3\left(2x-7\right)-x$
$\left(4a^2\right)+\left(7\right)^2$
$x+\frac{1}{x}=4$
$\left(3x^4y+12b^2\right)\left(12b^2-3x^4y\right)$
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